Question

A ball is thrown vertically upward with an initial velocity 38 m/s, simultaneously another ball is dropped from rest at a height of 152 m above the first ball. What will be the time of collision of these two balls -
2s
3 s
4 s
7.2s​

Answer 1

Answer:

Given Initial velocity of ball, u=49 m/s

Let the maximum height reached and time taken to reach that height be H and t respectively.

Assumption: g=9.8 m/s

2

holds true (maximum height reached is small compared to the radius of earth)

Velocity of the ball at maximum height is zero, v=0

v

2

−u

2

=2aH

0−(49)

2

=2×(−9.8)×H

⟹H=122.5 m

v=u+at

0=49−9.8t

⟹t=5 s

∴ Total time taken by ball to return to the surface, T=2t=10 s