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 upward with a velocity of $100m{s}^{-1}$. At what time will the bullet and the block meet ? Take g= $9.8m{s}^{-2}$.

• A. 4 s
• B. 3 s
• C. 2 s
• D. 1 s

Answer 1

The correct option is D

1 s

Let the distance covered by wooden block in downward direction before meeting bullet be $s_1$ and distance covered by bullet in upward direction before meeting wooden block be $s_2$
Given, $g=9.8m{s}^{-2}$
Let 't' be the time taken.
Height of the tower $=s_1+s_2=100m$.
For the block :
Initial velocity, u = 0
$\therefore$ $s_1=\frac{1}{2}g{t}^2=\frac{1}{2}\times 9.8\times t^2=4.9t^2$
For the bullet :
Initial velocity, $u=100m{s}^{-1}$
$\therefore$ $s_2=ut-\frac{1}{2}g{t}^2$
$=100t-\frac{1}{2}\times 9.8\times t^2=100t-4.9t^2$
Since, $s_1+s_2=100m$
$4.9t^2+100t-4.9t^2=100$
$\Rightarrow$ 100t = 100
$\Rightarrow$ t = 1 s.