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Question
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. 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 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?
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Solution 1
Initial velocity of the ball, u = 49 m/s
Acceleration, a = – g = – 9.8 m/s2
Case I:
When the lift was stationary, the boy throws the ball.
Taking upward motion of the ball,
Final velocity, v of the ball becomes zero at the highest point.
From first equation of motion, time of ascent (t) is given as:
`v = u+at`
`t = (v-u)/a`
=`(-49)/-9.8` = 5 s
But, the time of ascent is equal to the time of descent.
Hence, the total time taken by the ball to return to the boy’s hand = 5 + 5 = 10 s.
Case II:
The lift was moving up with a uniform velocity of 5 m/s. In this case, the relative velocity of the ball with respect to the boy remains the same i.e., 49 m/s. Therefore, in this case also, the ball will return back to the boy’s hand after 10 s.
Solution 2
When either the lift is at rest or the lift is moving either vertically upward or downward with a constant speed, we can apply three simple kinetnatic motion equations presuming a = ± g (as the case may be).In present case u = 49 ms-1 (upward) a = g = 9.8 ms-2(downward)
If the ball returns to boy’s hands after a time t, then displacement of ball relative to boy is zero i.e., s = 0. Hence, using equation s = ut + 1/2 at 2, we have
`0 = 49 +- -1/2xx 9.8 xx t^2`
`=>4.9 t^2 - 49t = 0 => t = 0 or 10 s`
As t = 0 is physically not possible, hence time t = 10s
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