A ball drops a man drops a ball downside from roof of the tower of height 400 metre at what time another ball is thrown upward with a velocity of 50 metre per second from the surface of the tower then they will meet at which height from the tower
Answers
Answered by
3
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
e.g collision occurs 80m height from the surface of tower .
Step-by-step explanation:
Similar questions