### Q.24:- 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 ?

**Answer:-**

Initial velocity of the ball, *u* = 49 m/s

Acceleration, *a* = – *g* = – 9.8 m/s^{2}

^{
}*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.