Q. A man drops a ball downside from the roof of a tower of height $400\ m$. At the same time another ball is thrown upside with a velocity $50\ m/s$. From the surface of the tower, then they will meet at which height from the surface of the tower
A) $100\ m$
B) $320\ m$
C) $80\ m$
D) $240\ m$
Explanation
(c) Let both balls meet at point $P$ after time $t$
The distance travelled by ball $A,\ h_{1} = \frac{1}{2}gt^{2}$
The distance travelled by ball $B,\ h_{2} = ut - \frac{1}{2}gt^{2}$
$h_{1} + h_{2} = 400\ m\ \Rightarrow ut = 400,\ t = 400/50 = 8\ sec$
$\therefore h_{1} = 320m$ and $h_{2} = 80\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.