**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:

Acceleration, a = - g = - 9.8 m/s2

Case I:

Final velocity, v of the ball becomes zero at
the highest point.

From first equation of motion, time of ascent (t) is given as:

From first equation of motion, time of ascent (t) is given as:

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.

## No comments:

## Post a Comment