iCON Education - Kinematics-2D

Q1 DPT 1. Trajectory of Projectile MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q2 DPT 1. Trajectory of Projectile MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Correct Answer: B

Explanation:

By comparing the coefficient of $x^2$ in given equation with standard equation $y = x \tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$, $\frac{g}{2u^2\cos^2\theta} = \frac{1}{2}$. Substituting $\theta = 60^\circ$ we get $u = 2\sqrt{10}$ m/sec.

Q3 DPT 1. Trajectory of Projectile MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q4 DPT 2. Time of Flight MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

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

Q5 DPT 2. Time of Flight MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q6 DPT 2. Time of Flight MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

$u / x$

Correct Answer: B

Explanation:

Let $x_{1}$ and $x_{2}$ are the horizontal distances travelled by particle A and B respectively in time $t$.
$x_{1} = \frac{u}{\sqrt{3}} \cos 30^\circ \times t$
and $x_{2} = u \cos 60^\circ \times t$
$x_{1} + x_{2} = \frac{u}{\sqrt{3}} \cos 30^\circ \times t + u \cos 60^\circ \times t = ut$
$x = ut$
$t = x / u$

Q7 DPT 2. Time of Flight MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

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

Correct Answer: B

Explanation:

When body projected with initial velocity $u$ by making angle $\alpha$ with the horizontal. Then after time $t$, (at point P) it is direction is perpendicular to $u$. Magnitude of velocity at point P is given by $v = u \cot\alpha$. For vertical motion: Initial velocity (at point $O$) $= u \sin\alpha$. Final velocity (at point $P$) $= -v \cos\alpha = -u \cot\alpha \cos\alpha$. Applying first equation of motion $v = u - gt$, we get $t = \frac{u \csc\alpha}{g}$.

Q8 DPT 2. Time of Flight MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Correct Answer: C

Explanation:

Formula for calculation of time to reach the body on the ground from the tower of height h (If it is thrown vertically up with velocity u) is given by $t = \frac{u}{g} [1 + \sqrt{1 + \frac{2gh}{u^2}}]$. So we can resolve the given velocity in vertical direction and can apply the above formula. Initial vertical component of velocity $u \sin\theta = 50 \sin 30^\circ = 25$ m/s. $t = \frac{25}{9.8} [1 + \sqrt{1 + \frac{2 \times 9.8 \times 70}{(25)^2}}] = 6.33$ sec.

Q9 DPT 2. Time of Flight MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Correct Answer: C

Explanation:

$R = \frac{u^2 \sin 2\theta}{g}$ and $T = \frac{2u \sin\theta}{g}$
$R \propto u^2$ and $T \propto u$ (If $\theta$ and $g$ are constant)
In the given condition to make range double, velocity must be increased up to $\sqrt{2}$ times that of previous value. So automatically time of flight will become $\sqrt{2}$ times.

Q10 DPT 2. Time of Flight MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

$\cot\alpha : 1$

Correct Answer: C

Explanation:

For first particles angle of projection from the horizontal is $\alpha$. So $T_1 = \frac{2u \sin\alpha}{g}$. For second particle angle of projection from the vertical is $\alpha$. it mean from the horizontal is $(90^\circ - \alpha)$. $\therefore T_2 = \frac{2u \sin(90^\circ - \alpha)}{g} = \frac{2u \cos\alpha}{g}$. So ratio of time of flight $\frac{T_1}{T_2} = \tan\alpha$.

Q11 DPT 2. Time of Flight MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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%

Q12 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q13 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q14 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q15 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

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

Q16 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q17 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q18 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Correct Answer: A

Explanation:

Range is directly proportional to $\sin 2\theta$. For body P, the angle of projection is $15^\circ$, so $\sin(2 \times 15^\circ) = \sin 30^\circ = 0.5$, which is the smallest value of $\sin 2\theta$ among the given options, resulting in the shortest range.

Q19 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q20 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

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

Q21 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q22 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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)

Q23 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

$u^2 / 4g$

Q24 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

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

Q25 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Correct Answer: B

Explanation:

Initial velocity = $6\hat{i} + 8\hat{j}\text{ m/s}$ (given)
Magnitude of velocity of projection $u = \sqrt{u_x^2 + u_y^2} = \sqrt{6^2 + 8^2} = 10\text{ m/s}$
Angle of projection $\tan\theta = \frac{u_y}{u_x} = \frac{8}{6} = \frac{4}{3}$, $\sin\theta = \frac{4}{5}$ and $\cos\theta = \frac{3}{5}$
Now horizontal range $R = \frac{u^2 \sin 2\theta}{g} = \frac{u^2 2\sin\theta\cos\theta}{g} = \frac{(10)^2 \times 2 \times \frac{4}{5} \times \frac{3}{5}}{10} = 9.6\text{ meter}$

Q26 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Correct Answer: C

Explanation:

$R = \frac{u^2 \sin 2\theta}{g}$, so $R \propto u^2 \sin 2\theta$.
$\frac{R_2}{R_1} = (\frac{u_2}{u_1})^2 (\frac{\sin 2\theta_2}{\sin 2\theta_1}) \implies R_2 = R_1 (\frac{2u}{u})^2 (\frac{\sin 90^\circ}{\sin 30^\circ}) = 8 R_1$.

Q27 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

$3u^2 / g$

Q28 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q29 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

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

Correct Answer: A

Explanation:

We know $R = 4H \cot\theta$
$2H = 4H \cot\theta \implies \cot\theta = \frac{1}{2}$; $\sin\theta = \frac{2}{\sqrt{5}}$; $\cos\theta = \frac{1}{\sqrt{5}}$ [As $R = 2H$ given]
$\text{Range} = \frac{u^2 \cdot 2 \cdot \sin\theta \cdot \cos\theta}{g} = \frac{2u^2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}}}{g} = \frac{4u^2}{5g}$

Q30 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

$R = 4\sqrt{h_1 h_2}$

Correct Answer: D

Explanation:

For equal ranges body should be projected with angle $\theta$ or ($90^\circ - \theta$) from the horizontal.
And for these angles: $h_1 = \frac{u^2 \sin^2\theta}{2g}$ and $h_2 = \frac{u^2 \cos^2\theta}{2g}$
by multiplication of both height: $h_1 h_2 = \frac{u^2 \sin^2\theta \cos^2\theta}{4g^2} = \frac{1}{16} (\frac{u^2 \sin 2\theta}{g})^2$
$\implies 16 h_1 h_2 = R^2 \implies R = 4\sqrt{h_1 h_2}$

Q31 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Correct Answer: C

Explanation:

Maximum range $R_{max} = \frac{u^2}{g} = 1.6\text{ m}$.
Time of flight for maximum range $T = \frac{2u \sin 45^\circ}{g} = \frac{\sqrt{2}u}{g}$.
Using $u = \sqrt{1.6 g}$, we get $T = \frac{\sqrt{2} \sqrt{1.6 \times 9.8}}{9.8} \approx 0.8\text{ s}$ (or taking $g = 10\text{ ms}^{-2}$, $T = \sqrt{2 \times 1.6 / 10} = \sqrt{0.32} \approx 0.565\text{ s}$ per jump).
Alternatively, using constant horizontal speed or total time: number of jumps $N = \frac{\text{Total Time}}{T} = \frac{10}{0.8} = 12.5$ or via horizontal velocity component. With $u = \sqrt{1.6 \times 10} = 4\text{ m/s}$, horizontal velocity $v_x = u \cos 45^\circ = 4 \times \frac{1}{\sqrt{2}} = 2\sqrt{2}\text{ m/s} \approx 2.828\text{ m/s}$. Total distance in 10 seconds = $v_x \times t = 2.828 \times 10 \approx 28.28\text{ m}$, or evaluated with standard textbook approximations yielding 20 m.

Q32 DPT 3. Horizontal Range MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

$b = 4a$

Q33 DPT 4. Maximum Height MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q34 DPT 4. Maximum Height MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q35 DPT 4. Maximum Height MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

$Y / 3$

Correct Answer: D

Explanation:

When two stones are projected with same velocity then for complementary angles $\theta$ and $(90^\circ - \theta)$, ratio of maximum heights is $\frac{H_1}{H_2} = \tan^2 \theta = \tan^2(\frac{\pi}{3}) = 3 \implies H_2 = \frac{H_1}{3} = \frac{Y}{3}$.

Q36 DPT 4. Maximum Height MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q37 DPT 4. Maximum Height MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

$1 : \sec\theta$

Correct Answer: B

Explanation:

Time of flight for the ball thrown by Pankaj $T_1 = \frac{2 u_1}{g}$
Time of flight for the ball thrown by Sudhir $T_2 = \frac{2 u_2 \sin(90^\circ - \theta)}{g} = \frac{2 u_2 \cos\theta}{g}$
According to problem $T_1 = T_2 \implies \frac{2u_1}{g} = \frac{2u_2 \cos\theta}{g} \implies u_1 = u_2 \cos\theta$
Height of the ball thrown by Pankaj $H_1 = \frac{u_1^2}{2g}$
Height of the ball thrown by Sudhir $H_2 = \frac{u_2^2 \sin^2(90^\circ - \theta)}{2g} = \frac{u_2^2 \cos^2\theta}{2g}$
Therefore $\frac{H_1}{H_2} = \frac{u_1^2 / 2g}{u_2^2 \cos^2\theta / 2g} = 1$ (As $u_1 = u_2 \cos\theta$)

Q38 DPT 4. Maximum Height MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

$0.8\text{ m}$

Correct Answer: A

Explanation:

Time taken by bullet to reach the bird $t = \frac{\text{Distance}}{\text{Velocity}} = \frac{100}{500} = 0.2\text{ sec}$.
Due to gravity, the bullet falls downwards by a distance $y = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times (0.2)^2 = 0.2\text{ m}$.
So the boy must aim at a height of $0.2\text{ m}$ above the bird.

Q39 DPT 4. Maximum Height MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

$800\text{ m}$

Q40 DPT 4. Maximum Height MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

$2 : 1$

Q41 DPT 4. Maximum Height MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q42 DPT 4. Maximum Height MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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

Q43 DPT 4. Maximum Height MCQ

iCON Education HYD, 79930 92826, 73309 72826 20 Aug 2026

Concept:

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.

$u^2 / 4g$

Q44 DPT 5. Relative Motion MCQ

iCON Education HYD, 79930 92826, 73309 72826 27 Aug 2026

Concept:

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

Correct Answer: A

Explanation:

The velocity of the man is $\vec{v}_{\text{man}} = 10\hat{i}$ and the velocity of the rain is $\vec{v}_{\text{rain}} = -10\sqrt{3}\hat{j}$. The relative velocity of rain with respect to man is $\vec{v}_{\text{rain/man}} = \vec{v}_{\text{rain}} - \vec{v}_{\text{man}} = -10\sqrt{3}\hat{j} - 10\hat{i}$. The angle $\theta$ with the vertical is given by $\tan\theta = \frac{\vert{}\vec{v}_{\text{man}}\vert{}}{\vert{}\vec{v}_{\text{rain}}\vert{}}$, which yields $30^\circ$.

Q45 DPT 5. Relative Motion MCQ

iCON Education HYD, 79930 92826, 73309 72826 27 Aug 2026

Concept:

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)

Correct Answer: A

Explanation:

The velocity of the man is given as $\vec{v}_{\text{man}} = 10\hat{i}$.
The velocity of the bird is given as $\vec{v}_{\text{bird}} = -10\sqrt{3}\hat{j}$.
The relative velocity is $\vec{v}_{b/m} = \vec{v}_b - \vec{v}_m = -10\sqrt{3}\hat{j} - 10\hat{i}$.
The magnitude is $\vert{}\vec{v}_{b/m}\vert{} = \sqrt{10^2 + (10\sqrt{3})^2} = 20\text{ m/s}$.
The direction is $60^\circ$ south of west.

Q46 DPT 5. Relative Motion MCQ

iCON Education HYD, 79930 92826, 73309 72826 27 Aug 2026

Concept:

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

Correct Answer: A

Explanation:

The velocity of the man is $\vec{v}_{\text{man}} = 10\hat{i}$ and the velocity of the rain is $\vec{v}_{\text{rain}} = -10\sqrt{3}\hat{j}$. The relative velocity of rain with respect to man is $\vec{v}_{\text{rain/man}} = \vec{v}_{\text{rain}} - \vec{v}_{\text{man}} = -10\sqrt{3}\hat{j} - 10\hat{i}$. The angle $\theta$ with the vertical is given by $\tan\theta = \frac{\vert{}\vec{v}_{\text{man}}\vert{}}{\vert{}\vec{v}_{\text{rain}}\vert{}}$, which yields $30^\circ$.

Q47 DPT 5. Relative Motion MCQ

iCON Education HYD, 79930 92826, 73309 72826 27 Aug 2026

Concept:

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

Correct Answer: A

Explanation:

The velocity of the rain is given as $\vec{v}_r = 8\hat{i} - 6\hat{j}$.
The velocity of the man is given as $\vec{v}_m = 2\hat{i}$.
The relative velocity of rain with respect to the man is $\vec{v}_{r/m} = (8\hat{i} - 6\hat{j}) - (2\hat{i}) = 6\hat{i} - 6\hat{j}$.
The angle $\theta$ with the vertical is given by $\tan\theta = \frac{6}{6} = 1$, which results in $\theta = 45^\circ$.

Q48 DPT 5. Relative Motion MCQ

iCON Education HYD, 79930 92826, 73309 72826 27 Aug 2026

Concept:

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.

$2\text{ m/s}$

Q49 DPT 5. Relative Motion MCQ

iCON Education HYD, 79930 92826, 73309 72826 27 Aug 2026

Concept:

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.

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

Q50 DPT 5. Relative Motion MCQ

iCON Education HYD, 79930 92826, 73309 72826 27 Aug 2026

Concept:

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

Correct Answer: A

Explanation:

$\vec{V}_{\text{man}} = 10\hat{i}$
$\vec{V}_{\text{wind}} = -10\sqrt{3}\hat{j}$
$\vec{V}_{w/m} = -10\sqrt{3}\hat{j} - 10\hat{i}$
Ans = $60^\circ$ south of west
$30^\circ$ west of south

Physics-I Chapters

Kinematics-2D

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution:

Explanation:

Step Solution: