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?
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. T
aking 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.
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