A wooden block of mass $20 g$ is dropped from the top of the cliff $50 m$ high. Simultaneously a bullet of mass $20 g$ is fired from the foot of the cliff upwards with a velocity $25 m s^{- 1}$ . The bullet and the wooden block will meet each other after time : ( $g = 10 m s^{- 2}$ )

2 s

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1 s

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0.5 s

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4 s

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Solution

The correct option is A $2 s$
Let they meet at a distance $h$ from the top and it takes time $t$ .
$- h = - \frac{1}{2} g t^{2}$
$h = 5 t^{2} \dots \dots (i)$
$50 - h = 25 t - \frac{1}{2} . g t^{2}$
from equation (1) -
$50 - 5 t^{2} = 25 t - 5 t^{2}$
$50 = 25 t$
$t = 2 sec$

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A wooden block is dropped from the top of a cliff 100 metre high and simultaneously a bullet of mass 10 gram is fired from the foot of the cliff upwards with a velocity of 100 m/s . The bullet and wooden block will meet each other after a time

Q. A wooden block of mass

20 g

is dropped from the top of the cliff

50 m

high. Simultaneously a bullet of mass

20 g

is fired from the foot of the cliff upwards with a velocity

25 m s^{- 1}

. The bullet and the wooden block will meet each other after time : (

g = 10 m s^{- 2}

)

Q. A wooden block of mass 10 gm is dropped from the top of a cliff 100 m high. Simultaneously a bullet of same mass is fired from the foot of the cliff vertically upwards with a velocity of

100 m s^{- 1}

. If the bullet after collision gets embedded in the block, the common velocity of the bullet and the block immediately after collision is

(g = 10 m s^{- 2})

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A wooden block of mass 20 g is dropped from the top of the cliff 50 m high. Simultaneously a bullet of mass 20 g is fired from the foot of the cliff upwards with a velocity 25 ms−1. The bullet and the wooden block will meet each other after time : (g=10 ms−2)

The correct option is A 2 sLet they meet at a distance h from the top and it takes time t. −h=−12gt2 h=5t2 ……(i) 50−h=25t−12.gt2 from equation (1) - 50−5t2=25t−5t2 50=25t t=2 sec

A wooden block of mass $20 g$ is dropped from the top of the cliff $50 m$ high. Simultaneously a bullet of mass $20 g$ is fired from the foot of the cliff upwards with a velocity $25 m s^{- 1}$ . The bullet and the wooden block will meet each other after time : ( $g = 10 m s^{- 2}$ )

The correct option is A $2 s$
Let they meet at a distance $h$ from the top and it takes time $t$ .
$- h = - \frac{1}{2} g t^{2}$
$h = 5 t^{2} \dots \dots (i)$
$50 - h = 25 t - \frac{1}{2} . g t^{2}$
from equation (1) -
$50 - 5 t^{2} = 25 t - 5 t^{2}$
$50 = 25 t$
$t = 2 sec$