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a boy standing on a stationary lift open from abo...

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?

Answer:

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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