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

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A body freely falling from the rest has a velocity ‘v’ after it falls through a height ‘h’. The distance it has to fall down for its velocity to become double, is

2 h

4 h

6 h

8 h

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Velocity of a body on reaching the point from which it was projected upwards, is

$v = 0$

$v = 2 u$

$v = 0 .5 u$

$v = - u$

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Explanation

Body reaches the point of projection with same speed in opposite direction.

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A body projected vertically upwards with a velocity u returns to the starting point in 4 seconds. If g = 10m/sec², the value of u is

5 m/sec

10 m/sec

15 m/sec

20 m/sec

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Explanation

Time of flight $T = \frac{2 u}{g} = 4 \sec \Rightarrow u = 20 m / s$

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Time taken by an object falling from rest to cover the height of h₁ and h₂ is respectively t₁ and t₂ . Then the ratio of t₁ to t₂ is

h₁ : h₂

$\sqrt{h_{1}} : \sqrt{h_{2}}$

h₁ : 2h₂

2h₁ : h₂

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Explanation

$t = \sqrt{\frac{2 h}{g}} \Rightarrow \frac{t_{1}}{t_{2}} = \sqrt{\frac{h_{1}}{h_{2}}}$

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A body is thrown vertically up from the ground. It reaches a maximum height of 100m in 5sec. After what time it will reach the ground from the maximum height position

1.2 sec

5 sec

10 sec

25 sec

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Explanation

Time of ascent = Time of descent = 5 sec

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A body thrown vertically upwards with an initial velocity u reaches maximum height in 6 seconds. The ratio of the distances travelled by the body in the first second and the seventh second is

1 : 1

11 : 1

1 : 2

1 : 11

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Explanation

Time of ascent $= \frac{u}{g} = 6 \sec \Rightarrow u = 60 m / s$

Distance in first second $h_{first} = 60 - \frac{g}{2} (2 \times 1 - 1) = 55 m$

Distance in seventh second will be equal to the distance in first second of vertical downward motion $h_{seventh} = \frac{g}{2} (2 \times 1 - 1) = 5 m$ ⇒ $h_{first} / h_{seventh} = 11 : 1$

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A particle is thrown vertically upwards. If its velocity at half of the maximum height is 10 m/s, then maximum height attained by it is (Take g = 10 m/s²)

8 m

10 m

12 m

16 m

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Explanation

Let particle thrown with velocity u and its maximum height is H then $H = \frac{u^{2}}{2 g}$

When particle is at a height H/2, then its speed is 10 m/s

From equation $v^{2} = u^{2} - 2 g h$

${(10)}^{2} = u^{2} - 2 g (\frac{H}{2}) = u^{2} - 2 g \frac{u^{2}}{4 g} \Rightarrow u^{2} = 200$

Maximum height $\Rightarrow H = \frac{u^{2}}{2 g} = \frac{200}{2 \times 10} = 10 m$

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Two balls A and B of same masses are thrown from the top of the building. A, thrown upward with velocity V and B, thrown downward with velocity V, then

Velocity of A is more than B at the ground

Velocity of B is more than A at the ground

Both A & B strike the ground with same velocity

None of these

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Explanation

$v^{2} = u^{2} + 2 g h \Rightarrow v = \sqrt{u^{2} + 2 g h}$

so for both the cases velocity will be equal.

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A ball is dropped from top of a tower of 100m height. Simultaneously another ball was thrown upward from bottom of the tower with a speed of 50 m/s ( $g = 10 m / s^{2}$ ). They will cross each other after

1 s

2 s

3 s

4 s

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A cricket ball is thrown up with a speed of 19.6 ms^–1. The maximum height it can reach is

9.8 m

19.6 m

29.4 m

39.2 m

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Explanation

$H_{\max} = \frac{u^{2}}{2 g} = \frac{19 .6 \times 19 .6}{2 \times 9 .8} = 19 .6 m$

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