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(B) When a stone is dropped from an ascending balloon,it inherits the velocity of the balloon at the moment of release. Let the height be $h$ and the

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Stone from $P$ reaches the ground first

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(B) When a stone is dropped from an ascending balloon,it inherits the velocity of the balloon at the moment of release. Let the height be $h$ and the acceleration due to gravity be $g$. The equation of motion for the stone is given by $h = -ut + \frac{1}{2}gt^2$,where $u$ is the upward velocity of the balloon.Rearranging this gives a quadratic equation in time $t$: $\frac{1}{2}gt^2 - ut - h = 0$.Solving for $t$ using the quadratic formula: $t = \frac{u + \sqrt{u^2 + 2gh}}{g}$.Since $g$ and $h$ are constant for all stones,the time $t$ depends on the initial upward velocity $u$. As $u$ increases,the time $t$ taken to reach the ground increases.Given velocities are $U_P = U$,$U_Q = 4U$,and $U_R = 8U$. Since $U_P Therefore,the stone from balloon $P$ reaches the ground first.

$P, Q$ and $R$ are three balloons ascending with velocities $U, 4U$ and $8U$ respectively. If stones of the same mass are dropped from each when they are at the same height $h$,which stone reaches the ground first?

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