Q251
Allen
JEE-Main Pattern
MCQ
Two balls are projected simultaneously from the top of a tall building, one vertically upward and other vertical downwards with the same speed of 60 m/s. Then time interval between the balls striking the ground is :-
A.
6 sec.
B.
10 sec.
C.
12 sec.
D.
9 sec.
Q252
Allen
JEE-Main Pattern
MCQ
A particle moves on x-axis as per equation $x = (t^{3} - 9t^{2} + 15t + 2)m$ . Distance travelled by the particle between t = 0 and t = 5 s is
A.
25 m
B.
39 m
C.
23 m
D.
52 m
Q253
Allen
JEE-Main Pattern
MCQ
The $\mathrm{XI^{th}}$ class students of ALLEN designed a rocket. The rocket was launched from cycle stand of ALLEN straight up into the air. At $t = 0$ , the rocket is at $y = 0$ with $V_{y}(t = 0) = 0$ . The velocity of the rocket is given by: $V_{y} = (24t - 3t^{2})m / s$ for $0 \leq t \leq t_{b}$ where $t_{b}$ is the time at which fuel burns out. Vertically upward direction is taken as positive. $(g = 10m / s^2)$ The time taken by rocket to reach its maximum height is
A.
4 sec.
B.
8 sec.
C.
8.8 sec.
D.
9.6 sec.
Q254
Allen
JEE-Main Pattern
MCQ
Acceleration-velocity graph of a moving particle is shown in figure. The particle is
A.
speeding up at P
B.
speeding up at Q
C.
speeding up at S
D.
speeding down at R
Q255
Allen
JEE-Main Pattern
MCQ
A body of mass 5 kg starts from the origin with an initial velocity $\vec{u} = (30\hat{i} + 40\hat{j})ms^{-1}$ . If a constant force $(-6\hat{i} - 5\hat{j})N$ acts on the body, the time in which the y component of the velocity becomes zero, is:
A.
5s
B.
20 s
C.
40 s
D.
80 s
Q256
Allen
JEE-Main Pattern
MCQ
A boy throws a ball from shoulder height at an initial velocity of 30 m/s. Spending 4.8 s in air, the ball is caught by another boy as the same shoulder-height level. What is the angle of projection?
A.
$37^{\circ}$
B.
$30^{\circ}$
C.
$53^{\circ}$
D.
$60^{\circ}$
Q257
Allen
JEE-Main Pattern
MCQ
A particle is projected from the ground with velocity u at angle $\theta$ with horizontal. The horizontal range, maximum height and time of flight are R, H and T respectively.
They are given by $R = \frac{u^{2} \sin 2\theta}{g} H = \frac{u^{2} \sin^{2} \theta}{2g}$ and $T = \frac{2u \sin \theta}{g}$ Now keeping u fixed, $\theta$ is varied from $30^{\circ}$ to $60^{\circ}$ , then
A.
R will first increase then decrease, H will increase and T will decrease
B.
R will first increase then decrease while H and T both will increase
C.
R will decrease while H and T both will increase
D.
R will increase while H and T both will also increase
Q258
Allen
JEE-Main Pattern
MCQ
Suppose a player hits several baseballs. Which baseball will be in the air for the longest time?
A.
The one with the farthest range.
B.
The one which reaches maximum height.
C.
The one with the greatest initial velocity.
D.
The one leaving the bat at $45^{\circ}$ with respect to the ground.
Q259
Allen
JEE-Main Pattern
MCQ
A ball is hit by a batsman at an angle of $37^{\circ}$ as shown in figure. The man standing at P should run at what minimum velocity so that he catches the ball before it strikes the ground. Assume that height of man is negligible in comparison to maximum height of projectile.
A.
3 m/s
B.
5 m/s
C.
9 m/s
D.
12 m/s
Q260
Allen
JEE-Main Pattern
MCQ
A particle P is projected from a point on the surface of smooth inclined plane (see figure). Simultaneously another particle Q is released on the smooth inclined plane from the same position. P and Q collide after t = 4 s. The speed of projection of P is :-
A.
5 m/s
B.
10 m/s
C.
15 m/s
D.
20 m/s
Q261
Allen
JEE-Main Pattern
MCQ
A trolley is moving horizontally with a constant velocity of v with respect to earth. A man starts running from one end of the trolley with a velocity 1.5 v with respect to trolley. After reaching the opposite end, the man returns back and continues running with a velocity of 1.5 v w.r.t. the trolley in the backward direction. If the length of the trolley is L then the displacement of the man with respect to earth during the process will be :-
A.
$2.5L$
B.
$1.5L$
C.
$\frac{5L}{3}$
D.
$\frac{4L}{3}$
Q262
Allen
JEE-Main Pattern
MCQ
An elevator car (lift) is moving upward with uniform acceleration of $2 \, m/s^{2}$ . At the instant, when its velocity is $2 \, m/s$ upwards a ball is thrown upward from its floor. The ball strikes back the floor $2 \, s$ after its projection. Find the velocity of projection of the ball relative to the lift.
A.
$10 \, m/s \uparrow$
B.
$10 \, m/s \downarrow$
C.
$12 \, m/s \uparrow$
D.
$12 \, m/s \downarrow$
Q263
Allen
JEE-Main Pattern
MCQ
A flag is mounted on a car moving due North with velocity of 20 km/hr. Strong winds are blowing due East with velocity of 20 km/hr. The flag will point in direction :-
A.
East
B.
North-East
C.
South-East
D.
South-West
Q264
Allen
JEE-Main Pattern
MCQ
Wind is blowing in the north direction at speed of 2 m/s which causes the rain to fall at some angle with the vertical. With what velocity should a cyclist drive so that the rain appears vertical to him:
A.
2 m/s south
B.
2 m/s north
C.
4 m/s west
D.
4 m/s south
Q265
Allen
JEE-Main Pattern
MCQ
A man is crossing a river flowing with velocity of 5 m/s. He reaches a point directly across the river at a distance of 60 m in 5 sec. His velocity in still water should be :-
A.
$12m / s$
B.
$13m / s$
C.
$5m / s$
D.
$10m / s$
Q266
Allen
NAT
Numerical
A rocket is fired vertically upwards with initial velocity 40 m/s at the ground level. Its engines then fired and it is accelerated at $2 \, m/s^{2}$ until it reaches an altitude of 1000 m. At that point the engines shut off and the rocket goes into free-fall. If the velocity (in m/s) just before it collides with the ground is $40\alpha$ . Then fill the value of $\alpha$ . Disregard air resistance ( $g = 10m/s^{2}$ ).
Correct Answer: 4
Explanation:
Ans. (4) $v^{2} = u^{2} + 2$ as $v^{2} = 40^{2} + 2(2)(1000) = 5600$ $v^{\prime 2} = v^{2} + 2gs$ $\Rightarrow v^{\prime 2} = 5600 + 2\times 10\times 100$ $\Rightarrow v^{\prime} = 160 = 40\alpha$ $\Rightarrow \alpha = 4$
Q267
Allen
NAT
Numerical
The vertical height y and horizontal distance x of a projectile on a certain planet are given by $x = (3t)m$ , $y = (4t - 6t^{2})m$ where t is in seconds. Find the speed of projection (in m/s).
Correct Answer: 5
Explanation:
Ans. (5) $v_{x} = \frac{dx}{dt} = 3, v_{y} = \frac{dy}{dt} = 4 - 12t$ At $t = 0, \vec{u} = u_{x}\hat{i} + u_{y}\hat{j} = 3\hat{i} + 4\hat{j}$ Speed of projection $= u = \sqrt{3^2 + 4^2} = 5m / s$
Q268
Allen
NAT
Numerical
A particle is thrown with a speed $60 \, ms^{-1}$ at an angle $60^{\circ}$ to the horizontal. When the particle makes an angle $30^{\circ}$ with the horizontal in downward direction, it's speed at that instant is v. What is the value of $v^{2}$ in SI units?
Correct Answer: 1200
Explanation:
Ans. (1200) Horizontal velocity remain constant. $60\cos 60^{\circ} = v\cos 30^{\circ}$ $v = 20\sqrt{3}$ $v^{2} = 1200$
Q269
Allen
NAT
Numerical
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high (in m) above the ground can the cricketer throw the same ball?
Correct Answer: 50
Explanation:
Ans. (50)
$H _ {\mathrm{max}} = u ^ {2} \sin^ {2} \theta / 2 g = u ^ {2} / 2 g$
$R _ {\max} = \left(u ^ {2} \sin 2 \theta / g\right) _ {\max} = u ^ {2} / g$
$2 H _ {\mathrm{max}} = \mathrm{Range}$
$2 H _ {\mathrm{max}} = 1 0 0$
$H _ {\mathrm{max}} = 5 0 m$
$H _ {\mathrm{max}} = u ^ {2} \sin^ {2} \theta / 2 g = u ^ {2} / 2 g$
$R _ {\max} = \left(u ^ {2} \sin 2 \theta / g\right) _ {\max} = u ^ {2} / g$
$2 H _ {\mathrm{max}} = \mathrm{Range}$
$2 H _ {\mathrm{max}} = 1 0 0$
$H _ {\mathrm{max}} = 5 0 m$
Q270
Allen
NAT
Numerical
A particle is projected upwards with a velocity of 100 m/s at an angle of $60^{\circ}$ with the vertical. Find the time (in sec) when the particle will move perpendicular to its initial direction, taking $g = 10 \, m/s^{2}$ .
Correct Answer: 20
Explanation:
Ans. (20) $U = 100\cos 30\hat{t} +100\sin 30^{\circ}\hat{j}$ $\vec{V} = 10\cos 30\hat{t} +(100\sin 30 - 10t)\hat{j}$ $\vec{U}.\vec{V} = 0$ $\Rightarrow \frac{100}{10\times\frac{1}{2}} = 20\sec$
Q271
Allen
NAT
Numerical
A ball is thrown horizontally from a cliff such that it strikes ground after 5 s. The line of sight from the point of projection to the point of landing makes an angle of $37^{\circ}$ with the horizontal. If the initial velocity of projection is $\frac{50x}{3}$ . Find x.
Correct Answer: 2
Explanation:
Ans. (2) $t^2 = \frac{2h}{g} = 25$ $\Rightarrow h = 125m$ $x = u_{x}\times t$ $h\cot 37^{\circ} = u_{x}\times 5$ $u_{x} = 100 / 3$
Q272
Allen
NAT
Numerical
A ball is dropped from rest from a tower of height 5m. As a result of the wind it lands at a distance 6m from the bottom of the tower as shown. Assuming no air resistance but that the wind gives the ball a constant horizontal velocity v. Find value of v in m/s.
Correct Answer: 6
Explanation:
Ans. (6)
$\begin{array}{l} {x = \sqrt {\frac {2 h}{g}} v} \\ {\Rightarrow \quad v = \frac {x}{\sqrt {\frac {2 h}{g}}} = \frac {6}{\sqrt {\frac {2 \times 5}{1 0}}} = 6 m / s} \end{array}$
$\begin{array}{l} {x = \sqrt {\frac {2 h}{g}} v} \\ {\Rightarrow \quad v = \frac {x}{\sqrt {\frac {2 h}{g}}} = \frac {6}{\sqrt {\frac {2 \times 5}{1 0}}} = 6 m / s} \end{array}$
Q273
Allen
NAT
Numerical
A cuboidal elevator cabin is shown in the figure. A ball is thrown from point A on the floor of cabin when the elevator is falling under gravity. The plane of motion is ABCD and the angle of projection of the ball with AB, relative to elevator, if the ball collides with point C, is $\alpha$ . Then find the value of tan $\alpha$ .
Correct Answer: 5
Explanation:
Ans. (5)
$\begin{array}{l} \tan \alpha = \frac {A D}{A B} \\ \frac {2 . 5}{0 . 5} \\ \tan \alpha = 5 \end{array}$
$\begin{array}{l} \tan \alpha = \frac {A D}{A B} \\ \frac {2 . 5}{0 . 5} \\ \tan \alpha = 5 \end{array}$
Q274
Allen
NAT
Numerical
Two particles P and Q are launched simultaneously as shown in figure. Find the minimum distance between particles in meters.
Correct Answer: 6
Explanation:
Ans. (6)
$\begin{array}{r l} & {\vec {v} _ {P Q} = \vec {v} _ {P} - \vec {v} _ {Q}} \\ & {\qquad = \left(\frac {1 0}{\sqrt {2}} \hat {\imath} + \frac {1 0}{\sqrt {2}} \hat {j}\right) - \left(- \frac {7 0}{\sqrt {2}} \hat {\imath} + \frac {7 0}{\sqrt {2}} \hat {j}\right)} \\ & {\qquad = 4 0 \sqrt {2} \hat {\imath} - 3 0 \sqrt {2} \hat {j}} \\ & {A B = 9 0 \times \frac {4}{3} = 1 2 0 m} \\ & {\Rightarrow \quad Q B = 1 0 m} \\ & {\Rightarrow \quad Q M = 1 0 \sin 3 7 ^ {\circ} = 1 0 \times \frac {3}{5} = 6 m} \end{array}$
$\begin{array}{r l} & {\vec {v} _ {P Q} = \vec {v} _ {P} - \vec {v} _ {Q}} \\ & {\qquad = \left(\frac {1 0}{\sqrt {2}} \hat {\imath} + \frac {1 0}{\sqrt {2}} \hat {j}\right) - \left(- \frac {7 0}{\sqrt {2}} \hat {\imath} + \frac {7 0}{\sqrt {2}} \hat {j}\right)} \\ & {\qquad = 4 0 \sqrt {2} \hat {\imath} - 3 0 \sqrt {2} \hat {j}} \\ & {A B = 9 0 \times \frac {4}{3} = 1 2 0 m} \\ & {\Rightarrow \quad Q B = 1 0 m} \\ & {\Rightarrow \quad Q M = 1 0 \sin 3 7 ^ {\circ} = 1 0 \times \frac {3}{5} = 6 m} \end{array}$
Q275
Allen
NAT
Numerical
Velocity of the boat with respect to river is 10 m/s. From point A it is steered in the direction shown. Find the distance (in m) it will reach on the opposite bank from point B.
Correct Answer: 30
Explanation:
Ans. (30)
$\begin{array}{l} {t = \frac {1 2 0}{8} = 1 5 \sec} \\ {\Delta = (6 - 4) \times 1 5 = 3 0 m} \end{array}$
$\begin{array}{l} {t = \frac {1 2 0}{8} = 1 5 \sec} \\ {\Delta = (6 - 4) \times 1 5 = 3 0 m} \end{array}$