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Motion in a Straight Line NEET Questions — Free MCQ Practice

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A particle is projected up with an initial velocity of $80 f t / \sec$ . The ball will be at a height of $96 f t$ from the ground after (Given $g = 32 f t / s^{2}$ )

A.

2.0 and 3.0 sec

B.

Only at 3.0 sec

C.

Only at 2.0 sec

D.

After 1 and 2 sec

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Explanation

$h = u t - \frac{1}{2} g t^{2} \Rightarrow 96 = 80 t - \frac{32}{2} t^{2}$

$\Rightarrow t^{2} - 5 t + 6 = 0 \Rightarrow t = 2$ sec or 3 sec

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A body falls from rest, its velocity at the end of first second is $(g = 32 f t / \sec^{2})$

A.

$16 f t / \sec$

B.

$32 f t / \sec$

C.

$64 f t / \sec$

D.

$24 f t / \sec$

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Explanation

$v = g \times t = 32 \times 1 = 32 f t / \sec$

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A stone thrown upward with a speed u from the top of the tower reaches the ground with a velocity 3u. The height of the tower is

A.

$3 u^{2} / g$

B.

$4 u^{2} / g$

C.

$6 u^{2} / g$

D.

$9 u^{2} / g$

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Explanation

$v^{2} = u^{2} + 2 g h \Rightarrow {(3 u)}^{2} = {(- u)}^{2} + 2 g h \Rightarrow h = \frac{4 u^{2}}{g}$

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Two stones of different masses are dropped simultaneously from the top of a building

A.

Smaller stone hit the ground earlier

B.

Larger stone hit the ground earlier

C.

Both stones reach the ground simultaneously

D.

Which of the stones reach the ground earlier depends on the composition of the stone

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Explanation

$t = \sqrt{\frac{2 h}{g}}$ and h and g are same.

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A body thrown up with an initial speed of $96 f t / \sec$ reaches the ground after $(g = 32 f t / \sec^{2})$

A.

3 sec

B.

6 sec

C.

12 sec

D.

8 sec

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Explanation

Time of flight $= \frac{2 u}{g} = \frac{2 \times 96}{32} = 6 s e c$

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A man in a balloon rising vertically with an acceleration of $4 .9 m / \sec^{2}$ releases a ball 2 sec after the balloon is let go from the ground. The greatest height above the ground reached by the ball is $(g = 9 .8 m / \sec^{2})$

A.

14.7 m

B.

19.6 m

C.

9.8 m

D.

24.5 m

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Explanation

Height travelled by ball (with balloon) in 2 sec

$h_{1} = \frac{1}{2} a t^{2} = \frac{1}{2} \times 4 .9 \times 2^{2} = 9 .8 m$

Velocity of the balloon after 2 sec

$v = a t = 4 .9 \times 2 = 9 .8 m / s$

Now if the ball is released from the balloon then it acquire same velocity in upward direction.

Let it move up to maximum height $h_{2}$

$v^{2} = u^{2} - 2 g h_{2}$ ⇒ $0 = {(9 .8)}^{2} - 2 \times (9 .8) \times h_{2}$

∴ $h_{2}$ = 4.9m

Greatest height above the ground reached by the ball $= h_{1} + h_{2} = 9 .8 + 4 .9 = 14 .7 m$

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A balloon is at a height of 81 m and is ascending upwards with a velocity of 12 m/s. A body of 2 kg weight is dropped from it. If $g = 10 m / s^{2}$ , the body will reach the surface of the earth in

A.

1.5 s

B.

4.025 s

C.

5.4 s

D.

6.75 s

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Explanation

$h = u t + \frac{1}{2} g t^{2} - 81 = 12 t - \frac{1}{2} \times 10 \times t^{2} \Rightarrow t = 5 .4 s e c$

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Water drops fall at regular intervals from a tap which is 5 m above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant

A.

2.50 m

B.

3.75 m

C.

4.00 m

D.

1.25 m

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Explanation

Time taken by first drop to reach the ground $t = \sqrt{\frac{2 h}{g}}$

$\Rightarrow t = \sqrt{\frac{2 \times 5}{10}} = 1 s e c$

As the water drops fall at regular intervals from a tap therefore time difference between any two drops $= \frac{1}{2} \sec$

In this given time, distance of second drop from the tap $= \frac{1}{2} g {(\frac{1}{2})}^{2} = \frac{5}{5} = 1 .25 m$

Its distance from the ground $= 5 - 1 .25 = 3 .75 m$

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A ball is thrown vertically upwards from the top of a tower at $4 .9 m s^{- 1}$ . It strikes the pond near the base of the tower after 3 seconds. The height of the tower is

A.

73.5 m

B.

44.1 m

C.

29.4 m

D.

None of these

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Explanation

$h = u t + \frac{1}{2} g t^{2}, t = 3 \sec, u = - 4 .9 m / s$

$\Rightarrow h = - 4 .9 \times 3 + 4 .9 \times 9 = 29 .4 m$

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A body starts to fall freely under gravity. The distances covered by it in first, second and third second are in ratio

A.

1 : 3 : 5

B.

1 : 2 : 3

C.

1 : 4 : 9

D.

1 : 5 : 6

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Explanation

$S_{n} = u + \frac{g}{2} (2 n - 1)$ ; when $u = 0$ , $S_{1} : S_{2} : S_{3} = 1 : 3 : 5$

Motion in a Straight Line MCQs for NEET — Physics Questions with Answers

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