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Two balls are dropped from heights h and 2h respectively from the earth surface. The ratio of time of these balls to reach the earth is

$1 : \sqrt{2}$

$\sqrt{2}$ : 1

2 : 1

1 : 4

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Explanation

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

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The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2m on the surface of A. What is the height of jump by the same person on the planet B

18m

$\frac{2}{3} m$

$\frac{2}{9} m$

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Explanation

$H_{\max} = \frac{u^{2}}{2 g} \Rightarrow H_{\max} \propto \frac{1}{g}$

On planet B value of g is $1 / 9$ times to that of A. So value of $H_{\max}$ will become 9 times i.e. $2 \times 9 = 18 m e t r e$

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A body falls from rest in the gravitational field of the earth. The distance travelled in the fifth second of its motion is $(g = 10 m / s^{2})$

25m

45m

90m

125m

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Explanation

$h_{n} = \frac{g}{2} (2 n - 1) \Rightarrow h_{5^{t h}} = \frac{10}{2} (2 \times 5 - 1) = 45 m$ .

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If a body is thrown up with the velocity of 15 m/s then maximum height attained by the body is (g = 10 m/s²)

11.25 m

16.2 m

24.5 m

7.62 m

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Explanation

$h_{\max} = \frac{u^{2}}{2 g} = \frac{{(15)}^{2}}{2 \times 10} = 11 .25 m$ .

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A balloon is rising vertically up with a velocity of 29 ms^–1. A stone is dropped from it and it reaches the ground in 10 seconds. The height of the balloon when the stone was dropped from it is (g = 9.8 ms^–2)

100 m

200 m

400 m

150 m

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Explanation

For stone to be dropped from rising balloon of velocity 29 m/s.

$u = - 29 m / s,$ t = 10 sec.

$∴ h = - 29 \times 10 + \frac{1}{2} \times 9 .8 \times 100$

= – 290 + 490 = 200 m.

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A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds

h/9 meters from the ground

7h/9 meters from the ground

8h/9 meters from the ground

17h/18 meters from the ground

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Two balls of same size but the density of one is greater than that of the other are dropped from the same height, then which ball will reach the earth first (air resistance is negligible)

Heavy ball

Light ball

Both simultaneously

Will depend upon the density of the balls

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Explanation

Since acceleration due to gravity is independent of mass, hence time is also independent of mass (or density) of object.

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A packet is dropped from a balloon which is going upwards with the velocity 12 m/s, the velocity of the packet after 2 seconds will be

–12 m/s

12 m/s

–7.6 m/s

7.6 m/s

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Explanation

When packet is released from the balloon, it acquires the velocity of balloon of value 12 m/s. Hence velocity of packet after 2 sec, will be

$v = u + g t = 12 - 9 .8 \times 2$ = – 7.6 m/s.

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If a freely falling body travels in the last second a distance equal to the distance travelled by it in the first three second, the time of the travel is

6 sec

5 sec

4 sec

3 sec

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Explanation

The distance traveled in last second.

$S_{Last} = u + \frac{g}{2} (2 t - 1) = \frac{1}{2} \times 9 .8 (2 t - 1) = 4 .9 (2 t - 1)$

and distance traveled in first three second,

$S_{Three} = 0 + \frac{1}{2} \times 9 .8 \times 9 = 44 .1 m$

According to problem $S_{Last} = S_{Three}$

$\Rightarrow 4 .9 (2 t - 1) = 44 .1 \Rightarrow 2 t - 1 = 9$ ⇒ t = 5 sec.

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The effective acceleration of a body, when thrown upwards with acceleration a will be :

$\sqrt{a - g^{2}}$

$\sqrt{a^{2} + g^{2}}$

$(a - g)$

$(a + g)$

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Explanation

Net acceleration of a body when thrown upward

= acceleration of body – acceleration due to gravity

= a – g

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