Question

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

• A. 4 s
• B. 3 s
• C. 2 s
• D. 1 s

Answer 1

The correct option is D 1 s

Let$s_1$ and $s_2$ be the displacements covered by wooden block and bullet respectively.
Given. height of the tower is 100 m
Therefore, $s=s_1+s_2=100m$.
From the second equation of motion,
$s=ut+\frac{1}{2}\times {at}^2$
Before the collision, in the case of block, initial velocity,u = 0 and the acceleration, a = g (acceleration due to gravity)
$s_1=\frac{1}{2}g{t}^2=\frac{1}{2}\times 9.8\times t^2=4.9t^2$
In the case of bullet, initial velocity, $u=100m{s}^{-1}$ and the acceleration, a = g (acceleration due to gravity)
$s_2=ut-\frac{1}{2}g{t}^2$
$=100t-\frac{1}{2}\times 9.8\times t^2=100t-4.9t^2$
Substituting the values in equation, $s_1+s_2=100m$, we get,
$4.9t^2+100t-4.9t^2=100$
$\Rightarrow$ 100t = 100
$\Rightarrow$ t = 1 s.