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 at a distance from top of the cliff : ( $g = 10 m s^{- 2}$ )

10 m

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20 m

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30 m

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40 m

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Solution

The correct option is B $20 m$
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) -
$50 - 5 t^{2} = 25 t - 5 t^{2}$
$50 = 25 t$
$t = 2 sec$
Putting this value of $t = 2 s$ in equation (i), we get
$h = 20 m$

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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 at a distance from top of the cliff : (g=10 ms−2)

The correct option is B 20 mLet 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 Putting this value of t=2 s in equation (i), we get h=20 m