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 m/s from the surface of the tower,find when and at which height from the surface of the tower the two balls meet together.

Answer 1

So,
400-x=50_/(x/5) - 5(_/(x/5))^2
=>400-x=10_/(5x) - x
=>400=10_/(5x)
SQUARING BOTH THE SIDES
=>160000=500x
=>x=1600/5=320 m
So, height of collision from ground or surface of tower is (400-320)=80 m

HOPE IT WILL HELP U

Answer 2

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 .