Question 1: Palette 1/63

DPT MCQ

The standard equation of trajectory for an oblique projectile is given by $y = x \tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$, where $\theta$ is the angle of projection and $u$ is the initial velocity. By comparing the coefficient of $x$ in the given equation with the standard equation, we can find the angle of projection $\theta$ using the relation $\tan\theta = \text{coefficient of } x$.

The trajectory of a projectile is represented by $y = \sqrt{3}x - gx^2/2$. The angle of projection is

A

$30^\circ$

B

$45^\circ$

C

$60^\circ$

D

None of these

Question 2: Palette 2/63

DPT MCQ

The standard equation of trajectory for an oblique projectile is given by $y = x \tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$, where $\theta$ is the angle of projection and $u$ is the initial velocity. By comparing the coefficient of $x^2$ in the given equation with the standard equation, we can find the initial velocity $u$ once the angle of projection $\theta$ is known.

The path followed by a body projected along y-axis is given as by $y = \sqrt{3}x - (1/2)x^2$, if $g = 10$ m/s, then the initial velocity of projectile will be (x and y are in m)

A

$3\sqrt{10}$ m/s

B

$2\sqrt{10}$ m/s

C

$10\sqrt{3}$ m/s

D

$10\sqrt{2}$ m/s

Question 3: Palette 3/63

DPT MCQ

The standard equation of trajectory in terms of horizontal range $R$ is given by $y = x \tan\theta [1 - \frac{x}{R}]$, where $R$ is the horizontal range of the projectile. By rewriting the given equation of trajectory in this standard form, the horizontal range $R$ can be directly identified.

The equation of projectile is $y = 16x - \frac{5x^2}{4}$. The horizontal range is

A

16 m

B

8 m

C

3.2 m

D

12.8 m

Question 4: Palette 4/63

DPT MCQ

For a given velocity, a projectile has the same horizontal range for two complementary angles of projection, say $\theta$ and $90^\circ - \theta$. The respective times of flight are given by $t_{1} = \frac{2u \sin\theta}{g}$ and $t_{2} = \frac{2u \cos\theta}{g}$. The product of these times of flight is proportional to the horizontal range $R$ through the relation $t_{1}t_{2} = \frac{2R}{g}$.

For a given velocity, a projectile has the same range R for two angles of projection if $t_{1}$ and $t_{2}$ are the times of flight in the two cases then

A

$t_{1}t_{2} \propto R^{2}$

B

$t_{1}t_{2} \propto R$

C

$t_{1}t_{2} \propto \frac{1}{R}$

D

$t_{1}t_{2} \propto \frac{1}{R^{2}}$

Question 5: Palette 5/63

DPT MCQ

The total time of flight $T$ for a projectile thrown with initial velocity $u$ at an angle $\theta$ with the horizontal is given by the formula $T = \frac{2u \sin\theta}{g}$, where $g$ is the acceleration due to gravity.

A body is thrown with a velocity of $9.8$ m/s making an angle of $30^\circ$ with the horizontal. It will hit the ground after a time

A

1.5 s

B

1 s

C

3 s

D

2 s

Question 6: Palette 6/63

DPT MCQ

The time after which the horizontal distance between the two particles moving towards each other becomes zero is calculated by equating the sum of the horizontal distances travelled by both particles to the total initial separation $x$, using the relation $x_{1} + x_{2} = x$ where horizontal distance is $x = (u \cos\theta) t$.

From the image given below, two particles are separated at a horizontal distance x as shown in figure. They are projected at the same time as shown in figure with different initial speed. The time after which the horizontal distance between the particles become zero is

A

$u / 2x$

B

$x / u$

C

$2u / x$

D

$u / x$

Question 7: Palette 7/63

DPT MCQ

When a projectile is thrown with an initial velocity $u$ at an angle $\alpha$ with the horizontal, its velocity at point $P$ where it becomes perpendicular to the initial direction is given by $v = u \cot\alpha$. By applying the first equation of motion along the vertical direction, the time taken $t$ to reach point $P$ is found using the formula $t = \frac{u \csc\alpha}{g}$.

A particle is projected from a point O with a velocity u in a direction making an angle $\alpha$ upward with the horizontal. After some time at point P it is moving at right angle with its initial direction of projection. The time of flight from O to P is

A

$\frac{u \sin\alpha}{g}$

B

$\frac{u \csc\alpha}{g}$

C

$\frac{u \tan\alpha}{g}$

D

$\frac{u \sec\alpha}{g}$

Question 8: Palette 8/63

DPT MCQ

The time taken by a ball thrown upwards from the top of a tower of height h to reach the ground is calculated using the formula $t = \frac{u_y}{g} [1 + \sqrt{1 + \frac{2gh}{u_y^2}}]$, where $u_y = u \sin\theta$ is the initial vertical component of the velocity, $g$ is the acceleration due to gravity, and $h$ is the height of the tower.

A ball is projected upwards from the top of tower with a velocity 50 ms-1 making angle 30 degrees with the horizontal. The height of the tower is 70 m. After how many seconds from the instant of throwing will the ball reach the ground

A

2.33 sec

B

5.33 sec

C

6.33 sec

D

9.33 sec

Question 9: Palette 9/63

DPT MCQ

The horizontal range of a projectile is given by $R = \frac{u^2 \sin 2\theta}{g}$ and the time of flight is given by $T = \frac{2u \sin\theta}{g}$. For a constant angle of projection $\theta$ and acceleration due to gravity $g$, the range $R$ is proportional to $u^2$ ($R \propto u^2$) and the time of flight is proportional to the initial velocity $u$ ($T \propto u$).

If for a given angle of projection, the horizontal range is doubled, the time of flight becomes

A

4 times

B

2 times

C

$\sqrt{2}$ times

D

$1/\sqrt{2}$ times

Question 10: Palette 10/63

DPT MCQ

The time of flight for a projectile is given by the formula $T = \frac{2u \sin\theta}{g}$, where $\theta$ is the angle of projection with the horizontal. For the second particle projected at an angle $\alpha$ from the vertical, its angle of projection with the horizontal is $90^\circ - \alpha$, so its time of flight is given by $T_2 = \frac{2u \cos\alpha}{g}$. The ratio of their times of flight is found by taking $T_1 / T_2$.

A particle is thrown with velocity u at an angle $\alpha$ from the horizontal. Another particle is thrown with the same velocity at an angle $\alpha$ from the vertical. The ratio of times of flight of two particles will be

A

$\tan 2\alpha : 1$

B

$\cot 2\alpha : 1$

C

$\tan\alpha : 1$

D

$\cot\alpha : 1$

Question 11: Palette 11/63

DPT MCQ

The time of flight of a projectile is inversely proportional to the effective acceleration acting in the downward direction ($T = \frac{2u \sin\theta}{g}$). When air friction adds an upward retardation of $g/10$, the effective acceleration increases to $g_1 = g + g/10$, leading to a decreased time of flight $T_2$ compared to the initial time $T_1$.

The friction of the air causes vertical retardation equal to one tenth of the acceleration due to gravity (Take g = 10 ms-2). The time of flight will be decreased by

A

0%

B

1%

C

9%

D

11%

Question 12: Palette 12/63

DPT MCQ

The horizontal range of a projectile launched from a height and returning to the same vertical level is given by the formula $R = \frac{u^2 \sin 2\theta}{g}$, where $u$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.

A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at an angle of 30 degrees with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground (g = 10 m/s^2, sin 30 degrees = 1/2, cos 30 degrees = \sqrt{3}/2)

A

8.66 m

B

5.20 m

C

4.33 m

D

2.60 m

Question 13: Palette 13/63

DPT MCQ

The horizontal distance covered by a projectile (such as an athlete in a long-jump event) is governed by the horizontal range formula $R = \frac{u^2 \sin 2\theta}{g}$, which depends directly on the initial speed $u$ and the angle of projection $\theta$.

Which of the following sets of factors will affect the horizontal distance covered by an athlete in a long-jump event

A

Speed before he jumps and his weight

B

The direction in which he leaps and the initial speed

C

The force with which he pushes the ground and his speed

D

The direction in which he leaps and the weight

Question 14: Palette 14/63

DPT MCQ

The maximum height $H$ reached by a projectile is given by $H = \frac{u^2 \sin^2\theta}{2g}$ and the time of flight is given by $T = \frac{2u \sin\theta}{g}$. By taking the ratio of maximum height to the square of the time of flight ($H / T^2$), we find that the terms involving initial velocity and angle cancel out, yielding a constant ratio dependent only on acceleration due to gravity $g$.

For a projectile, the ratio of maximum height reached to the square of flight time is ($g = 10$ ms-2)

A

5 : 4

B

5 : 2

C

5 : 1

D

10 : 1

Question 15: Palette 15/63

DPT MCQ

The maximum horizontal range for a projectile is achieved when the angle of projection is $45^\circ$, and it is given by the formula $R_{max} = \frac{u^2}{g}$, where $u$ is the initial speed and $g$ is the acceleration due to gravity. By knowing $R_{max}$ and $g$, the throwing speed $u$ can be found using the relation $u = \sqrt{R_{max} g}$.

A cricketer can throw a ball to a maximum horizontal distance of 100 m. The speed with which he throws the ball is (to the nearest integer)

A

$30\text{ ms}^{-1}$

B

$42\text{ ms}^{-1}$

C

$32\text{ ms}^{-1}$

D

$35\text{ ms}^{-1}$

Question 16: Palette 16/63

DPT MCQ

When two bodies are projected with the same initial velocity at complementary angles ($\theta$ and $90^\circ - \theta$, such as $30^\circ$ and $60^\circ$), their horizontal ranges are equal because the range formula $R = \frac{u^2 \sin 2\theta}{g}$ yields the same value for both angles since $\sin(2 \times 30^\circ) = \sin(2 \times 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$.

If two bodies are projected at $30^\circ$ and $60^\circ$ respectively, with the same velocity, then

A

Their ranges are same

B

Their heights are same

C

Their times of flight are same

D

All of these

Question 17: Palette 17/63

DPT MCQ

The horizontal range of a projectile is directly proportional to the horizontal component of its initial velocity $u_x = u \cos\theta$ and the time of flight $T$, such that $R = u_x T$. Since the horizontal component of velocity remains constant throughout the flight, the path with the largest horizontal range possesses the highest initial horizontal velocity component.

From the image given below, figure shows four paths for a kicked football. Ignoring the effects of air on the flight, rank the paths according to initial horizontal velocity component, highest first

A

1, 2, 3, 4

B

2, 3, 4, 1

C

3, 4, 1, 2

D

4, 3, 2, 1

Question 18: Palette 18/63

DPT MCQ

The horizontal range of a projectile is given by the formula $R = \frac{u^2 \sin 2\theta}{g}$. For a given initial velocity $u$, the horizontal range depends directly on $\sin 2\theta$. The value of $\sin 2\theta$ increases as the angle increases from $0^\circ$ to $45^\circ$ and decreases symmetrically from $45^\circ$ to $90^\circ$, meaning angles closest to $45^\circ$ yield the largest range, while angles furthest from $45^\circ$ (closest to $0^\circ$ or $90^\circ$) yield the shortest range.

Four bodies P, Q, R and S are projected with equal velocities having angles of projection $15^\circ$, $30^\circ$, $45^\circ$ and $60^\circ$ with the horizontal respectively. The body having shortest range is

A

P

B

Q

C

R

D

S

Question 19: Palette 19/63

DPT MCQ

The horizontal range of a projectile is given by the formula $R = \frac{u^2 \sin 2\theta}{g}$. For a given angle of projection and surface conditions, the horizontal range is directly proportional to the square of the initial speed, meaning $R \propto u^2$.

A particle covers 50 m distance when projected with an initial speed. On the same surface it will cover a distance, when projected with double the initial speed

A

100 m

B

150 m

C

200 m

D

250 m

Question 20: Palette 20/63

DPT MCQ

The horizontal range of a projectile is calculated using the formula $R = \frac{u^2 \sin 2\theta}{g}$, where $u$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.

A bullet is fired from a canon with velocity 500 m/s. If the angle of projection is $15^\circ$ and $g = 10\text{ m/s}^2$. Then the range is

A

$25 \times 10^3\text{ m}$

B

$12.5 \times 10^3\text{ m}$

C

$50 \times 10^2\text{ m}$

D

$25 \times 10^2\text{ m}$

Question 21: Palette 21/63

DPT MCQ

The horizontal range of a projectile is given by the formula $R = \frac{u^2 \sin 2\theta}{g}$. For a given speed $v$ and angle $\theta$, the range is inversely proportional to the acceleration due to gravity $g$ ($R \propto 1/g$). Since the acceleration due to gravity on the moon is one-sixth of that on earth, the range on the moon will be 6 times the range on earth.

A projectile thrown with a speed $v$ at an angle $\theta$ has a range $R$ on the surface of earth. For same $v$ and $\theta$, its range on the surface of moon will be

A

R / 6

B

6 R

C

R / 36

D

36 R

Question 22: Palette 22/63

DPT MCQ

For maximum horizontal range, the angle of projection is $\theta = 45^\circ$. The speed of a projectile is minimum at the highest point of its trajectory where it possesses only the horizontal component of velocity. The co-ordinates of this highest point in terms of horizontal range $R$ and maximum height $H$ are $x = R / 2$ and $y = H = R / 4$.

A projectile is thrown into space so as to have maximum horizontal range R. Taking the point of projection as origin, the co-ordinates of the point where the speed of the particle is minimum are

A

(R, R)

B

(R, R / 2)

C

(R / 2, R / 4)

D

(R, R / 4)

Question 23: Palette 23/63

DPT MCQ

At the highest point of a projectile's trajectory, the velocity consists solely of the horizontal component, which is given by $u \cos\theta$. Given that this speed is equal to $1/\sqrt{2}$ of the initial speed $u$, we can find the angle of projection $\theta$. Then, the horizontal range can be calculated using the range formula $R = \frac{u^2 \sin 2\theta}{g}$.

The speed of a projectile at the highest point becomes $1/\sqrt{2}$ times its initial speed. The horizontal range of the projectile will be

A

$u^2 / g$

B

$u^2 / 2g$

C

$u^2 / \sqrt{2}g$

D

$u^2 / 4g$

Question 24: Palette 24/63

DPT MCQ

When bullets are fired in all directions with the same initial speed $u$, they can reach a maximum horizontal distance equal to the maximum range of a projectile, which is given by $R_{max} = \frac{u^2}{g}$ for $\theta = 45^\circ$. Thus, they cover a circular area on the ground with a radius equal to this maximum range.

A large number of bullets are fired in all directions with same speed u. What is the maximum area on the ground on which these bullets will spread

A

$\pi \frac{u^2}{g}$

B

$\pi \frac{u^4}{g^2}$

C

$\pi^2 \frac{u^4}{g^2}$

D

$\pi^2 \frac{u^2}{g^2}$

Question 25: Palette 25/63

DPT MCQ

The horizontal range of a projectile is calculated using the formula $R = \frac{u^2 \sin 2\theta}{g} = \frac{2u_x u_y}{g}$, where $u_x$ and $u_y$ are the horizontal and vertical components of the initial velocity vector, and $g$ is the acceleration due to gravity.

A projectile is projected with initial velocity $6\hat{i} + 8\hat{j}\text{ m/sec}$. If $g = 10\text{ ms}^{-2}$, then horizontal range is

A

4.8 metre

B

9.6 metre

C

19.2 metre

D

14.0 metre

Question 26: Palette 26/63

DPT MCQ

The horizontal range of a projectile is given by the formula $R = \frac{u^2 \sin 2\theta}{g}$, which shows that the range is directly proportional to the square of the initial velocity and the sine of twice the angle of projection ($R \propto u^2 \sin 2\theta$).

A projectile thrown with an initial speed u and angle of projection 15 degrees to the horizontal has a range R. If the same projectile is thrown at an angle of 45 degrees to the horizontal with speed 2u, its range will be

A

12 R

B

3 R

C

8 R

D

4 R

Question 27: Palette 27/63

DPT MCQ

At the highest point of a projectile trajectory, the velocity of the particle consists only of its horizontal component, which is given by $u \cos\theta$. Given that this velocity is half of the initial velocity $u$, we can determine the angle of projection $\theta$, and then calculate the horizontal range using the range formula $R = \frac{u^2 \sin 2\theta}{g}$.

The velocity at the maximum height of a projectile is half of its initial velocity of projection u. Its range on the horizontal plane is

A

$\sqrt{3}u^2 / 2g$

B

$u^2 / 3g$

C

$3u^2 / 2g$

D

$3u^2 / g$

Question 28: Palette 28/63

DPT MCQ

The horizontal range of a projectile is given by the formula $R = \frac{2 u_x u_y}{g}$, where $u_x$ and $u_y$ are the horizontal and vertical components of the initial velocity, respectively, and $g$ is the acceleration due to gravity.

A projectile is thrown from a point in a horizontal place such that its horizontal and vertical velocity component are 9.8 m/s and 19.6 m/s respectively. Its horizontal range is

A

4.9 m

B

9.8 m

C

19.6 m

D

39.2 m

Question 29: Palette 29/63

DPT MCQ

The relation between the horizontal range $R$ and the maximum height $H$ of a projectile is given by $R = 4H \cot\theta$, where $\theta$ is the angle of projection. Using the given condition $R = 2H$, we can find the trigonometric values for the angle of projection and calculate the range in terms of $v$ and $g$.

A particle is projected with a velocity v such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g is acceleration due to gravity)

A

$4v^2 / 5g$

B

$4g / 5v^2$

C

$v^2 / g$

D

$4v^2 / \sqrt{5}g$

Question 30: Palette 30/63

DPT MCQ

For complementary angles of projection $\theta$ and $90^\circ - \theta$, the horizontal range $R$ remains the same. The maximum heights attained for these respective angles are given by $h_1 = \frac{u^2 \sin^2\theta}{2g}$ and $h_2 = \frac{u^2 \cos^2\theta}{2g}$, and their product relates to the range through the formula $R = 4\sqrt{h_1 h_2}$.

The range $R$ of projectile is same when its maximum heights are $h_1$ and $h_2$. What is the relation between $R$ and $h_1$ and $h_2$

A

$R = \sqrt{h_1 h_2}$

B

$R = \sqrt{2h_1 h_2}$

C

$R = 2\sqrt{h_1 h_2}$

D

$R = 4\sqrt{h_1 h_2}$

Question 31: Palette 31/63

DPT MCQ

The maximum horizontal range of a projectile is achieved at an angle of $45^\circ$ and is given by $R_{max} = \frac{u^2}{g} = 1.6\text{ m}$. The time of flight for this maximum range is $T = \frac{2u \sin 45^\circ}{g} = \frac{\sqrt{2}u}{g}$. The total horizontal distance covered in multiple jumps is the product of the range per jump and the number of jumps made in the given total time.

A grasshopper can jump maximum distance 1.6 m. It spends negligible time on the ground. How far can it go in 10 seconds

A

5 m

B

10 m

C

20 m

D

40 m

Question 32: Palette 32/63

DPT MCQ

The angle of projection $\theta$ can be found using the components of the initial velocity where $\tan\theta = \frac{b}{a}$. The relation between horizontal range $R$ and maximum height $H$ is given by $R = 4H \cot\theta$. Using the given condition $R = 2H$, we can relate $\tan\theta$ and $\cot\theta$ to find the relationship between $a$ and $b$.

A projectile is thrown with an initial velocity of $v = a\hat{i} + b\hat{j}$, if the range of projectile is double the maximum height reached by it then

A

$a = 2b$

B

$b = a$

C

$b = 2a$

D

$b = 4a$

Question 33: Palette 33/63

DPT MCQ

The maximum horizontal range of a projectile is given by $R_{max} = \frac{u^2}{g}$, which occurs at an angle of projection $\theta = 45^\circ$. The maximum height attained when thrown vertically upwards (at $\theta = 90^\circ$) is given by $H_{max} = \frac{u^2}{2g}$.

A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same effort, he throws the ball vertically upwards. The maximum height attained by the ball is

A

100 m

B

80 m

C

60 m

D

50 m

Question 34: Palette 34/63

DPT MCQ

The time of flight for a projectile is given by the formula $T = \frac{2u \sin \theta}{g}$, which allows us to find the vertical component of the initial velocity $u \sin \theta$. The maximum height attained above the point of projection is calculated using the formula $H = \frac{(u \sin \theta)^2}{2g}$.

A ball thrown by one player reaches the other in 2 sec. the maximum height attained by the ball above the point of projection will be about

A

10 m

B

7.5 m

C

5 m

D

2.5 m

Question 35: Palette 35/63

DPT MCQ

The maximum height of a projectile is given by the formula $H = \frac{u^2 \sin^2 \theta}{2g}$. For a given initial velocity $u$, the ratio of maximum heights for two different angles of projection $\theta_1$ and $\theta_2$ is given by $\frac{H_1}{H_2} = \frac{\sin^2 \theta_1}{\sin^2 \theta_2} = \tan^2 \theta$ when the angles are complementary, or calculated directly using their respective sine squared values.

Two stones are projected with the same magnitude of velocity, but making different angles with horizontal. The angle of projection of one is $\pi / 3$ and its maximum height is $Y$, the maximum height attained by the other stone with as $\pi / 6$ angle of projection is

A

$Y$

B

$2 Y$

C

$3 Y$

D

$Y / 3$

Question 36: Palette 36/63

DPT MCQ

The maximum height of a projectile is given by the formula $H = \frac{u^2 \sin^2 \theta}{2g}$. For a constant angle of projection $\theta$, the maximum height is directly proportional to the square of the initial velocity ($H \propto u^2$).

If the initial velocity of a projectile be doubled. Keeping the angle of projection same, the maximum height reached by it will

A

Remain the same

B

Be doubled

C

Be quadrupled

D

Be halved

Question 37: Palette 37/63

DPT MCQ

The time of flight for a projectile or vertically thrown object is given by $T = \frac{2 u_y}{g}$. The maximum height attained is given by $H = \frac{u_y^2}{2g}$. When two bodies have the same time of flight, their vertical components of initial velocity are equal, leading to a relation between their initial velocities and their respective maximum heights.

From the image given below, Pankaj and Sudhir are playing with two different balls of masses m and 2m respectively. If Pankaj throws his ball vertically up and Sudhir at an angle $\theta$, both of them stay in our view for the same period. The height attained by the two balls are in the ratio

A

$2 : 1$

B

$1 : 1$

C

$1 : \cos\theta$

D

$1 : \sec\theta$

Question 38: Palette 38/63

DPT MCQ

When a projectile is aimed directly at a target, gravity causes it to fall below the target by a vertical distance of $h = \frac{1}{2}gt^2$. The time $t$ to reach the horizontal distance $x$ is determined by the constant horizontal component of the velocity, $t = \frac{x}{u \cos\theta}$.

A boy aims a gun at a bird from a point, at a horizontal distance of $100\text{ m}$. If the gun can impart a velocity of $500\text{ ms}^{-1}$ to the bullet. At what height above the bird must he aim his gun in order to hit it (take $g = 10\text{ ms}^{-2}$)

A

$0.2\text{ m}$

B

$0.4\text{ m}$

C

$0.6\text{ m}$

D

$0.8\text{ m}$

Question 39: Palette 39/63

DPT MCQ

The maximum horizontal range of a projectile occurs at an angle of $45^\circ$ and is given by $R_{max} = \frac{u^2}{g} = 400\text{ m}$. The relation between the maximum horizontal range and maximum height $H$ is given by $R = 4H \cot\theta$. For $45^\circ$, this simplifies to $R = 4H$, from which the maximum height can be calculated.

The maximum horizontal range of a projectile is $400\text{ m}$. The maximum height attained by it will be

A

$100\text{ m}$

B

$200\text{ m}$

C

$400\text{ m}$

D

$800\text{ m}$

Question 40: Palette 40/63

DPT MCQ

The maximum height of a projectile is given by the formula $H = \frac{u^2 \sin^2 \theta}{2g}$. For bodies projected with the same initial velocity $u$, the ratio of their maximum heights is proportional to the square of the sine of their respective projection angles: $\frac{H_1}{H_2} = \frac{\sin^2 \theta_1}{\sin^2 \theta_2}$.

Two bodies are projected with the same velocity. If one is projected at an angle of $30^\circ$ and the other at an angle of $60^\circ$ to the horizontal, the ratio of the maximum heights reached is

A

$3 : 1$

B

$1 : 3$

C

$1 : 2$

D

$2 : 1$

Question 41: Palette 41/63

DPT MCQ

The time of flight of a projectile is given by $T = \frac{2u \sin \theta}{g}$, which allows us to find the vertical component of the initial velocity $u \sin \theta = \frac{Tg}{2}$. The maximum height attained is given by the formula $H = \frac{(u \sin \theta)^2}{2g}$.

If time of flight of a projectile is 10 seconds. Range is 500 m. The maximum height attained by it will be

A

125 m

B

50 m

C

100 m

D

150 m

Question 42: Palette 42/63

DPT MCQ

The maximum horizontal range of a projectile is achieved at an angle of $45^\circ$ and is given by $R_{max} = \frac{u^2}{g} = 80\text{ m}$. The maximum height attained when projected vertically upward is given by $H_{max} = \frac{u^2}{2g}$.

A man can throw a stone 80 m. The maximum height to which he can raise the stone is

A

10 m

B

15 m

C

30 m

D

40 m

Question 43: Palette 43/63

DPT MCQ

For complementary angles of projection $\theta$ and $90^\circ - \theta$, the horizontal range of a projectile remains the same. The maximum heights attained for these complementary angles are given by $y_1 = \frac{u^2 \sin^2\theta}{2g}$ and $y_2 = \frac{u^2 \cos^2\theta}{2g}$. Their sum can be found by adding these two expressions together using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$.

A ball is thrown at different angles with the same speed u and from the same points and it has same range in both the cases. If y1 and y2 be the heights attained in the two cases, then y1 + y2 =

A

$u^2 / g$

B

$2u^2 / g$

C

$u^2 / 2g$

D

$u^2 / 4g$

Question 44: Palette 44/63

DPT MCQ

The relative velocity of rain with respect to man is given by the vector subtraction $\vec{v}_{r/m} = \vec{v}_r - \vec{v}_m$. The direction in which the man must hold his umbrella is parallel to the relative velocity vector of the rain with respect to the man. If $\theta$ is the angle made with the vertical, then $\tan\theta = \frac{v_m}{v_r}$.

A man is moving along +x-axis (east) with speed $10\text{ m/s}$ & rain is falling vertically downward with speed $10\sqrt{3}\text{ m/s}$. In which direction man should hold umbrella to protect himself.

A

$30^\circ$ with vertical towards east

B

$60^\circ$ with vertical towards east

C

$30^\circ$ with horizontal towards east

D

$45^\circ$ with vertical towards east

Question 45: Palette 45/63

DPT MCQ

The relative velocity of an object B with respect to object A is given by the vector subtraction formula $\vec{v}_{b/m} = \vec{v}_b - \vec{v}_m$. The magnitude is calculated as $\vert{}\vec{v}_{b/m}\vert{} = \sqrt{v_x^2 + v_y^2}$ and the direction is given by $\tan\theta = \frac{v_{\text{man}}}{v_{\text{bird}}}$.

A man is moving in east direction with speed $10\text{ m/s}$ in a car and a bird is flying with speed $10\sqrt{3}\text{ m/s}$ in south direction. Find velocity of bird observed by man.

A

$20\text{ m/s}$ ($60^\circ$ south of west)

B

$20\text{ m/s}$ ($30^\circ$ south of west)

C

$10\sqrt{3}\text{ m/s}$ ($45^\circ$ south of west)

D

$10\text{ m/s}$ ($60^\circ$ south of west)

Question 46: Palette 46/63

DPT MCQ

The relative velocity of rain with respect to man is given by the vector subtraction $\vec{v}_{r/m} = \vec{v}_r - \vec{v}_m$. The direction in which the man must hold his umbrella is parallel to the relative velocity vector of the rain with respect to the man. If $\theta$ is the angle made with the vertical, then $\tan\theta = \frac{v_m}{v_r}$.

A man is moving along +x-axis (east) with speed $10\text{ m/s}$ & rain is falling vertically downward with speed $10\sqrt{3}\text{ m/s}$. In which direction man should hold umbrella to protect himself.

A

$30^\circ$ with vertical towards east

B

$60^\circ$ with vertical towards east

C

$30^\circ$ with horizontal towards east

D

$45^\circ$ with vertical towards east

Question 47: Palette 47/63

DPT MCQ

The relative velocity of rain with respect to the man is calculated using the vector difference $\vec{v}_{r/m} = \vec{v}_r - \vec{v}_m$. The umbrella must be held parallel to the direction of this relative velocity vector to shield against the incoming rain.

Rain is falling with a speed of $10\text{ m/s}$ at an angle of $53^\circ$ with the vertical, and a man is moving along the east direction with a speed of $2\text{ m/s}$. In which direction should the man hold his umbrella to protect himself?

A

$45^\circ$ with the vertical towards the east

B

$30^\circ$ with the vertical towards the east

C

$60^\circ$ with the vertical towards the east

D

$37^\circ$ with the vertical towards the east

Question 48: Palette 48/63

DPT MCQ

The relative velocity of rain with respect to the man is given by $\vec{v}_{r/m} = \vec{v}_r - \vec{v}_m$. For the rain to appear falling vertically downward, the horizontal component of the relative velocity of rain must be zero, meaning the velocity of the man must equal the horizontal component of the rain's velocity.

Rain is falling with a speed of $10\text{ m/s}$ at an angle of $53^\circ$ with the vertical, and a man is moving along the east direction with a speed of $2\text{ m/s}$. What should be the velocity of the man so that the rain appears to be falling vertically downward to him?

A

$6\text{ m/s}$

B

$8\text{ m/s}$

C

$10\text{ m/s}$

D

$2\text{ m/s}$

Question 49: Palette 49/63

DPT MCQ

The relative velocity of rain with respect to the moving man makes the horizontal component of the rain's velocity equal to the man's velocity so that the rain appears to fall vertically. The relation is given by $v \sin 30^\circ = v_{\text{man}}$, where $v$ is the actual speed of the rain.

To a stationary man, rain appear to be falling at an angle $30^\circ$ with the vertical. As he start moving with speed of $5\text{ m/s}$ he feels that rain is falling vertically. Find speed of rain.

A

$10\text{ m/s}$

B

$5\text{ m/s}$

C

$20\text{ m/s}$

D

$10\sqrt{3}\text{ m/s}$

Question 50: Palette 50/63

DPT MCQ

The flag flutters in the direction of the relative velocity of the wind with respect to the man, which is given by the formula $\vec{v}_{w/m} = \vec{v}_w - \vec{v}_m$. The direction is determined by finding the angle made by the relative velocity vector with the coordinate axes using trigonometry.

A man holding a flag is running with speed $10\text{ m/s}$ along east. If wind speed is $10\sqrt{3}\text{ m/s}$ along south then in which direction flag will flutter?

A

$60^\circ$ south of west

B

$30^\circ$ south of west

C

$45^\circ$ south of west

D

$60^\circ$ north of east

Question 51: Palette 51/63

DPT MCQ

The motion can be analyzed easily in the non-inertial frame of reference of the lift (w.r.t lift). The relative acceleration of the coin with respect to the lift is given by $\vec{a}_{\text{coin/lift}} = \vec{a}_{\text{coin}} - \vec{a}_{\text{lift}}$, where the acceleration of the coin under gravity is $-g$ ($10\text{ m/s}^2$ downwards). The kinematic equation used is $s = ut + \frac{1}{2}at^2$.

From the image given below, a lift of height $32\text{ m}$ is going up with acceleration $a = 6\text{ m/s}^2$. When the velocity of the lift is $15\text{ m/s}$ upward, a coin is dropped from the ceiling at $t = 0$. Find the time when the coin will hit the floor.

A

$4\text{ sec}$

B

$2\text{ sec}$

C

$3\text{ sec}$

D

$1\text{ sec}$

Question 52: Palette 52/63

DPT MCQ

The motion is analyzed from the non-inertial frame of reference of the accelerating lift. The relative acceleration of the coin with respect to the lift is given by $\vec{a}_{\text{coin/lift}} = \vec{a}_{\text{coin}} - \vec{a}_{\text{lift}}$, where acceleration due to gravity is taken as $-10\text{ m/s}^2$ downwards. The kinematic equation used is $s = ut + \frac{1}{2}at^2$.

From the image given below, at $t = 0$ a coin is dropped from the ceiling of a lift of height $63\text{ m}$ which is moving upwards with an acceleration $a = 4\text{ m/s}^2$ and velocity $v = 19.7\text{ m/s}$. Find the time when the coin will hit the floor.

A

$3\text{ sec}$

B

$4\text{ sec}$

C

$2\text{ sec}$

D

$5\text{ sec}$

Question 53: Palette 53/63

DPT MCQ

The relative motion of the ball with respect to the lift is analyzed using the relative acceleration formula $\vec{a}_{\text{ball/lift}} = \vec{a}_{\text{ball}} - \vec{a}_{\text{lift}}$. The kinematic equation used is $s = ut + \frac{1}{2}at^2$.

From the image given below, a ball is thrown downward at $10\text{ m/s}$ with respect to a lift of height $100\text{ m}$ which is moving with an acceleration $a = 2\text{ m/s}^2$ and velocity $v = 40\text{ m/s}$. When will the coin strike the floor?

A

$3\text{ sec}$

B

$4\text{ sec}$

C

$5\text{ sec}$

D

$2\text{ sec}$

Question 54: Palette 54/63

DPT MCQ

The relative motion of the coin dropped from the ceiling is analyzed in the non-inertial frame of reference of the lift using the relative acceleration formula $\vec{a}_{\text{coin/lift}} = \vec{a}_{\text{coin}} - \vec{a}_{\text{lift}}$. The kinematic equation used is $s = ut + \frac{1}{2}at^2$.

From the image given below, when will the coin strike the floor of the lift, given its height is $120\text{ m}$, acceleration $a = 5\text{ m/s}^2$ upwards, and initial velocity $v = 10\text{ m/s}$ upwards?

A

$3\text{ sec}$

B

$5\text{ sec}$

C

$4\text{ sec}$

D

$2\text{ sec}$

Question 55: Palette 55/63

DPT MCQ

The relative motion of the coin dropped from the ceiling is analyzed in the non-inertial frame of reference of the lift using the relative acceleration formula $\vec{a}_{\text{coin/lift}} = \vec{a}_{\text{coin}} - \vec{a}_{\text{lift}}$. The kinematic equation used is $s = ut + \frac{1}{2}at^2$.

From the image given below, when will the coin strike the floor of the lift, given its height is $120\text{ m}$, acceleration $a = 5\text{ m/s}^2$ upwards, and initial velocity $v = 10\text{ m/s}$ upwards?

A

$3\text{ sec}$

B

$5\text{ sec}$

C

$4\text{ sec}$

D

$2\text{ sec}$

Question 56: Palette 56/63

DPT MCQ

The motion of a boat across a flowing river involves relative velocity. The resultant velocity of the boat ($v_{\text{res}}$) is the vector sum of the velocity of the boat in still water ($v_{\text{boat}}$) and the velocity of the river ($v_{\text{river}}$). Assuming they are perpendicular, the relation is given by:
$v_{\text{res}}^2 = v_{\text{boat}}^2 + v_{\text{river}}^2$

A boat is sent across a river with a velocity of $8\text{ km/hr}$. If the resultant velocity of boat is $10\text{ km/hr}$, then velocity of the river is :

A

$10\text{ km/hr}$

B

$8\text{ km/hr}$

C

$6\text{ km/hr}$

D

$4\text{ km/hr}$

Question 57: Palette 57/63

DPT MCQ

The relative velocity of an object A with respect to object B is given by the vector difference between their velocities: $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$.

A boat is moving with velocity of $3\hat{i} + 4\hat{j}$ in river (wrt ground) and water is moving with a velocity of $-3\hat{i} - 4\hat{j}$ with respect to ground. Relative velocity of boat with respect to water is :

A

$-6\hat{i} - 8\hat{j}$

B

$6\hat{i} + 8\hat{j}$

C

$8\hat{i}$

D

$6\hat{i}$

Question 58: Palette 58/63

DPT MCQ

The average speed of a swimmer crossing a river is defined as the total distance traveled divided by the total time taken. For crossing in the shortest possible time, the swimmer swims perpendicular to the river flow direction. The average speed formula is given by:
$\text{Average Speed} = \frac{\sqrt{v_{\text{swimmer}}^2 + v_{\text{river}}^2}}{2}$

A swimmer can swim in still water with a speed of $\sqrt{5}\text{ m/s}$. While crossing a river his average speed is $3\text{ m/s}$. If he cross the river in the shortest possible time, what is the speed of flow of water?

A

$1\text{ m/s}$

B

$2\text{ m/s}$

C

$3\text{ m/s}$

D

$4\text{ m/s}$

Question 59: Palette 59/63

DPT MCQ

For a swimmer to take the shortest possible path (straight across, perpendicular to the flow), the resultant velocity vector must be directed straight across the river (due South). The relation between the river velocity $v_r$, the swimmer's velocity $v_s$, and the angle $\theta$ with the downstream is given by:
$\sin(\theta) = \frac{v_r}{v_s}$

A river is flowing from W to E with a speed of $5\text{ m/min}$. A man can swim in still water with a velocity $10\text{ m/min}$. In which direction should the man swim so as to take the shortest possible path to go to the south.

A

$30^\circ$ with downstream

B

$60^\circ$ with downstream

C

$120^\circ$ with downstream

D

South

Question 60: Palette 60/63

DPT MCQ

When a swimmer wants to cross a river to reach a point directly opposite (minimum drift), the component of his velocity along the river flow must cancel the velocity of the river. The angle $\theta$ with the normal to the river flow is given by:
$\sin(\theta) = \frac{v_{\text{river}}}{v_{\text{swimmer}}}$

A man can swim in still water at $6\text{ km/h}$. He want to reach a point just opposite his starting point. In which direction he should swim, if the river is flowing at $3\text{ km/h}$ ? from the image given below

A

$30^\circ$ against the current from normal to the direction of flow

B

$45^\circ$ against the current from normal to the direction of flow

C

$60^\circ$ against the current from normal to the direction of flow

D

$90^\circ$ against the current from normal to the direction of flow

Question 61: Palette 61/63

DPT MCQ

The time taken $t$ to cross a river of width $d$ when swimming at right angles (perpendicular) to the river flow is determined solely by the width of the river and the component of the swimmer's velocity perpendicular to the flow ($v_{\text{swimmer}}$). The formula is given by:
$t = \frac{d}{v_{\text{swimmer}}}$

A man can swim in still water at $6\text{ km/h}$. He want to reach a point just opposite his starting point. In which direction he should swim, if the river is flowing at $3\text{ km/h}$ ? from the image given below Calculate the time taken by the man to cross the river at right angles to the flow; if the width of the river is $\sqrt{3}\text{ km}$.

A

$10\text{ minutes}$

B

$20\text{ minutes}$

C

$30\text{ minutes}$

D

$40\text{ minutes}$

Question 62: Palette 62/63

DPT MCQ

For minimum distance path (shortest path), the man must swim upstream at an angle $\theta$ with the line perpendicular to the river flow such that the component of his swimming velocity opposing the river flow cancels the river velocity $u$. The condition is given by $v \sin\theta = u$, where $v$ is the swimming velocity of the man and $u$ is the river velocity. The minimum distance covered is equal to the width of the river $d$, and the time taken to cross is $t = \frac{d}{v \cos\theta}$.

From the image given below, a river is flowing with velocity $5\text{ m/s}$ wrt ground. A man can swim with velocity $10\text{ m/s}$ wrt river. The width of the river is $100\text{ m}$. In which direction man should swim so that he crosses that river in min distance travel also find time taken to cross the river.

A

$30^\circ$ upstream, $11.55\text{ s}$

B

$30^\circ$ downstream, $10\text{ s}$

C

$60^\circ$ upstream, $12\text{ s}$

D

$45^\circ$ upstream, $14.14\text{ s}$

Question 63: Palette 63/63

DPT MCQ

The time taken to cross a river of width $d$ when swimming at an angle $\theta$ to the river flow is given by $t = \frac{d}{v \sin\theta}$, where $v$ is the velocity of the man relative to the river. The total horizontal drift along the river flow is given by $\text{Drift} = (u + v \cos\theta) t$, where $u$ is the river velocity.

From the image given below, in the following figure if a man is swimming with a velocity $10\text{ m/s}$ with respect to the river at an angle of $37^\circ$ to the river velocity, find the time taken to cross the river and the drift if the velocity of the river is $10\text{ m/s}$ and the width of the river is $30\text{ m}$.

A

$4\text{ s}, 80\text{ m}$

B

$5\text{ s}, 90\text{ m}$

C

$6\text{ s}, 100\text{ m}$

D

$5\text{ s}, 75\text{ m}$