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Question

(C) $1$. First,find the velocity of the helicopter at height $h$. Since it starts from rest with acceleration $g$,$v^2 = u^2 + 2as \Rightarrow v^2 = 0

Vedclass · Accepted Answer

$t = 3.4 \sqrt{\frac{h}{g}}$

Vedclass · Answer

(C) $1$. First,find the velocity of the helicopter at height $h$. Since it starts from rest with acceleration $g$,$v^2 = u^2 + 2as \Rightarrow v^2 = 0 + 2gh \Rightarrow v = \sqrt{2gh}$.$2$. When the packet is dropped,it has an initial upward velocity $u = \sqrt{2gh}$.$3$. The packet moves upward,reaches a maximum height,and then falls to the ground. The maximum height reached by the packet from the point of release is $H_{max} = \frac{u^2}{2g} = \frac{2gh}{2g} = h$.$4$. The total height of the packet from the ground is $H_{total} = h + h = 2h$.$5$. Using the equation of motion $s = ut + \frac{1}{2}at^2$ for the downward journey (taking upward as positive,$s = -h$,$u = \sqrt{2gh}$,$a = -g$):$-h = \sqrt{2gh} \cdot t - \frac{1}{2}gt^2$$\frac{1}{2}gt^2 - \sqrt{2gh} \cdot t - h = 0$Multiplying by $2/g$: $t^2 - 2\sqrt{\frac{2h}{g}} \cdot t - \frac{2h}{g} = 0$.$6$. Solving for $t$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$t = \frac{2\sqrt{2h/g} + \sqrt{8h/g + 8h/g}}{2} = \sqrt{\frac{2h}{g}} + \sqrt{\frac{4h}{g}} = (\sqrt{2} + 2) \sqrt{\frac{h}{g}} \approx (1.414 + 2) \sqrt{\frac{h}{g}} = 3.414 \sqrt{\frac{h}{g}}$.

$A$ helicopter rises from rest on the ground vertically upwards with a constant acceleration $g$. $A$ food packet is dropped from the helicopter when it is at a height $h$. The time taken by the packet to reach the ground is close to [$g$ is the acceleration due to gravity].

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