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(A) $1$. Velocity of the balloon at $t = 8\, s$: $v = a 	imes t = 1.25 	imes 8 = 10\, m/s$ (upward).$2$. Height of the balloon at $t = 8\, s$: $s =

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Reach the ground in $4\, s$

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(A) $1$. Velocity of the balloon at $t = 8\, s$: $v = a 	imes t = 1.25 	imes 8 = 10\, m/s$ (upward).$2$. Height of the balloon at $t = 8\, s$: $s = \frac{1}{2} 	imes a 	imes t^2 = \frac{1}{2} 	imes 1.25 	imes 8^2 = 40\, m$.$3$. When the stone is released,its initial velocity is $u = 10\, m/s$ (upward) and its initial position is $h = 40\, m$ above the ground.$4$. Using the equation of motion for displacement: $s = ut + \frac{1}{2}at^2$. Here,$s = -40\, m$ (downward),$u = 10\, m/s$,$a = -g = -10\, m/s^2$.$-40 = 10t - 5t^2 \implies 5t^2 - 10t - 40 = 0 \implies t^2 - 2t - 8 = 0$.Solving the quadratic equation: $(t - 4)(t + 2) = 0$. Since $t > 0$,$t = 4\, s$.$5$. The displacement of the stone from the release point to the ground is $40\, m$ (downward).$6$. The stone moves upward until its velocity becomes zero: $v^2 - u^2 = 2as \implies 0 - 10^2 = 2(-10)d \implies d = 5\, m$.$7$. Total distance covered = $40\, m$ (upward) + $5\, m$ (upward) + $5\, m$ (downward) = $50\, m$.

$A$ balloon starts rising from the ground with an acceleration of $1.25\, m/s^2$. After $8\,s$,a stone is released from the balloon. The stone will $(g = 10\, m/s^2)$:

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