Question

A man drops a ball downside from the roof of a tower of height 400 meters. At the same time another ball is thrown upside with a velocity $50\frac{meter}{sec}$. from the surface of the tower, then they will meet at which height from the surface of the tower

• A. 100 meters
• B. 320 meters
• C. 80 meters
• D. 240 meters

Answer 1

The correct option is C 80 meters

Let both balls meet at point P after time t.​

The distance travelled by ball A, $h_1=\frac{1}{2}g{t}^2$
The distance travelled by ball B, $h_2=ut-\frac{1}{2}g{t}^2$
$h_1+h_2=400m\Rightarrow ut=400,t=\frac{400}{50}=8sec$
$\therefore$ $h_1=320m$ and $h_2=80m$