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A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum inttial 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?
Solution
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
$v=u+a t$
$t=\frac{v-u}{a}$
$=\frac{-49}{-9.8}=5 s$
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 $10\;s.$
