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 50m/s from the surface of the tower,the

Answer 1

A ball dropped from the roof of height 400m Let after t time it meets with another ball .
displacement covered by ball during this time ( S1 ) = ut + 1/2at² 
= 0× t + 1/2 × 10 × t² 
= 5t² 

another ball is thrown from the surface of tower with speed 50m/sec .
then, displacement covered by another ball (S2) = ut +1/2at² 
= 50t - 1/2× 10t² 
=50t -5t² 

for collision, 
S1 + S2 = 400 
5t² + 50t - 5t² = 400 
50t = 400 
t = 8sec 

hence, body collide in 8sec.

then, displacement covered by first body during this time= 5 × 64 = 320m 

e.g collision occurs 80m height from the surface of tower .