A boy standing on a stationary lift (open from above) throws a ball upwards with
the maximum initial speed he can, equal to 49 m s–1. How much time does the ball
take to return to his hands? If the lift starts moving up with a uniform speed of
5 m s-1 and the boy again throws the ball up with the maximum speed he can, how
long does the ball take to return to his hands ?
Answers
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Initial velocity = u = 49m/s
Acceleration = a = -g = -9.8m/s²
When the ball reaches the top, velocity = 0
Hence,
v = u + at
0 = 49 + -9.8t
t = 49 / 9.8 = 5s
Hence ball reaches the top in 5s and comes down in 5s
Therefore the total time taken by the ball to reach back = 5+5 = 10s
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10 Seconds
In this case it does not matter whether whether the lift is moving vertically upward or downward or it is at rest at a constant speed.
Given:
Initial Velocity (u) = 49 m s^-1
g = 9.8 m/s^2
Distance = 0
(Because if the ball happens to returns to the boy’s hands after a time t, then displacement of ball relative to boy )
Here we use the kinematic equation:
S = ut + 1/2 at^2
Substituting the values known to us in this kinematic equation we get:
0 = 49 + / - 1/2 x 9.8 x t^2
- 49 +/- 4.9 x t^2
4.9t^2 - 49t = 0
t = 0
(This has to be rejected because time cannot be 0 seconds as it is not possible)
Time = 10 seconds
Therefore the time the ball takes to return to his hands is 10 seconds.