Question

From the top of the tower of height $400m$, a ball is dropped by a man, simultaneously from the base of the tower, another ball is thrown up with a velocity $50m/s$; at what distance (in $m$) will they meet from the base of the tower?

Answer 1

Step 1: Draw a rough diagram of the given situation.

Step 2: Find the distance travelled by both the balls.
Formula Used: $s=ut+\frac{1}{2}g{t}^2$
Since both the balls start from rest therefore initial velocity is zero i.e. $u=0$.
Let, both balls meet at point $P$ after time $t$.
The distance travelled by ball $A$,
$h_1=\frac{1}{2}g{t}^2\dots \left(i\right)$
The distance travelled by ball $B$,
$h_2=ut-\frac{1}{2}g{t}^2\dots \left(ii\right)$

Step 3: Find the value of $h_1$ and $h_2$.
Since, $h_1+h_2=400m$
From equation $\left(i\right)$ and $\left(ii\right)$
$h_1+h_2=ut=400$
$t=\frac{400}{50}=8\text{sec}$
$\therefore h_1=320m$ and $h_2=80m$
Hence, balls will meet from the base of the tower at a height of $80m$.

Final answer: $80m$.