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+4 votes
in Physics by (2.8k points)
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

1 Answer

+3 votes
by (29.7k points)
Best answer

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 .

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Thank you, it was really helpful

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