Question

From the top of tower of height 100 m a 10 gm block is dropped freely and a 6 gm bullet is fired vertically upward from the foot of the tower with velocity 100 $m{s}^{-1}$ simultaneously. They collide and stick together after time. Find the time after which they collide with each other.

Answer 1

In such situation, we must take a proper coordinate system and sign convention.

Consider foot of tower as $y=0$ and upwards as +ve

acceleration due to gravity acts downwards.

$g=10m/s^2$

Given: $|{\mu }_1|=|{\mu }_2|=100m/s$.

We get the following equations of motion:

$y_1={\mu }_{1}t-\frac{1}{2}g{t}^2$ (for bullet)

$y_2=100-{\mu }_{2}t-\frac{1}{2}g{t}^2$ (for block)

When they collide, they will same position $(y_1=y_2)$ time $t$, So:-

$y_1=y_2$

$\Rightarrow {\mu }_{1}t-\frac{1}{2}g{t}^2=100-{\mu }_{2}t-\frac{1}{2}g{t}^2$

$\Rightarrow ({\mu }_1+{\mu }_2)t=100$

$\Rightarrow t=\frac{100}{{\mu }_1+{\mu }_2}=\frac{100}{100+100}=0.5s$.

$\therefore$ They collide at $t=0.5s$