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(A) $1$. Initial motion of the balloon: $u = 0$,$a = 4 \ ft/sec^2$,$t = 5 \ sec$. Height of the balloon at $t = 5 \ sec$: $h_0 = ut + \frac{1}{2}at^2

Vedclass · Accepted Answer

$T = 5/2 \ sec$

Vedclass · Answer

(A) $1$. Initial motion of the balloon: $u = 0$,$a = 4 \ ft/sec^2$,$t = 5 \ sec$. Height of the balloon at $t = 5 \ sec$: $h_0 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} 	imes 4 	imes 5^2 = 50 \ ft$. Velocity of the balloon at $t = 5 \ sec$: $v_0 = u + at = 0 + 4 	imes 5 = 20 \ ft/sec$. $2$. Motion of the stone after being dropped: The stone has an initial upward velocity $v_0 = 20 \ ft/sec$ and acceleration $g = -32 \ ft/sec^2$. Using $s = ut + \frac{1}{2}at^2$ for the stone to reach the ground $(s = -50 \ ft)$: $-50 = 20T + \frac{1}{2}(-32)T^2$ $-50 = 20T - 16T^2$ $\Rightarrow 16T^2 - 20T - 50 = 0$ $\Rightarrow 8T^2 - 10T - 25 = 0$. Solving for $T$: $T = \frac{10 \pm \sqrt{100 - 4(8)(-25)}}{2(8)} = \frac{10 \pm \sqrt{900}}{16} = \frac{10 \pm 30}{16}$. Since $T > 0$,$T = \frac{40}{16} = 2.5 \ sec = 5/2 \ sec$. $3$. Height of the balloon when the stone reaches the ground: The balloon continues to ascend with $a = 4 \ ft/sec^2$ for an additional $T = 2.5 \ sec$. $H = h_0 + v_0 T + \frac{1}{2}aT^2 = 50 + 20(2.5) + \frac{1}{2}(4)(2.5)^2 = 50 + 50 + 12.5 = 112.5 \ ft$.

$A$ balloon starting from rest is ascending from the ground with a uniform acceleration of $4 \ ft/sec^2$. At the end of $5 \ sec$,a stone is dropped from it. If $T$ is the time taken for the stone to reach the ground and $H$ is the height of the balloon when the stone reaches the ground,then:

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