Question

A wooden block of mass $10g$ is dropped from rest from the top of a cliff $100m$ high. Simultaneously, a bullet of mass $10g$ is fired from the foot of the cliffupward with a velocity of $100m{s}^{-1}$. After what time, will the bullet and the block meet?

• A. $4s$
• B. $3s$
• C. $2s$
• D. $1s$

Answer 1

The correct option is D $1s$

Height of the tower is:
$H=100m=|s_1|+|s_2|$
where:
$|s_1|:$ Distance of the wooden block
$|s_2|:$ Distance of the bullet

Consider the downward direction to be positive.
For both, block and bullet, acceleration is equal to the acceleration due to gravity.
$a_1=a_2=g\approx 10m/s^2$

Using equation of motion for the block:
$s_1=u_{1}t+\frac{1}{2}{a}_1{t}^2$
$u_1=0$ because the block falls from rest.
$|s_1|=\frac{1}{2}g{t}^2$

Using equation of motion for the bullet:
$s_2=u_{2}t+\frac{1}{2}{a}_2{t}^2$

Note that upward direction is negative. So, displacement, and initial velocity are positive.
$-|s_2|=-|u_2|t+\frac{1}{2}g{t}^2$
$|s_2|=|u_2|t-\frac{1}{2}g{t}^2$

Now,
$|s_1|+|s_2|=H$
$\frac{1}{2}g{t}^2+|u_2|t-\frac{1}{2}g{t}^2=H$
$⟹t=\frac{H}{|u_2|}$
$t=\frac{100m}{100m/s}=1s$