two balls are dropped from 1 km . second ball is dropped after 5 sec . find the time after which they cross each other. Assume that the ball landing on the ground immediately bounces back with same velocity with no loss of energy.
Answers
Answer:
16.78 sec
Explanation:
two balls are dropped from 1 km
First ball will reach at ground in T sec
Using S = ut + (1/2)at²
S = 1 km = 1000m
u = 0 (dropped)
a = g = 9.8 m/s²
1000 = 0 + (1/2)(9.8) t²
=> t² = 10000/49
=> t = 100/7 sec
Velocity after 100/7 sec
= 0 + 9.8(100/7)
= 140 m/s
Velocity of second ball after t-5 sec = 100/7 - 5 = 65/7 sec
= 0 + 9.8(65/7)
= 91 m/s
Distance covered by second ball = 91²/2*9.8 = 422.5 m
Now Both balls has to cover 1000 - 422.5 = 577.5m
Let say t sec more are taken
Distance covered by first Ball
= 140t - (1/2)(9.8)t²
Distance covered by Second Ball
= 91t + (1/2)(9.8)t²
=> Distance covered by both balls = 140t + 91t = 231t
231t = 577.5
=> t = 17.5/7
Total Time taken from drop of 1st ball
= 100/7 + 17.5/7
= 117.5/7
= 16.78 sec