Question in the description:

Question

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 50m/s, from the surface bottom. 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

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 

collision occurs 80m height from the surface of tower .

Comments

Thank you, it was really helpful