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Question

(C) The time taken by the food packet to reach the ground is given by $t = \sqrt{\frac{2h}{g}}$.The horizontal distance (range) covered by the packet

Vedclass · Accepted Answer

$\sqrt{\frac{2v^{2}h}{g}+h^{2}}$

Vedclass · Answer

(C) The time taken by the food packet to reach the ground is given by $t = \sqrt{\frac{2h}{g}}$.The horizontal distance (range) covered by the packet during this time is $R = v \cdot t = v \sqrt{\frac{2h}{g}}$.When the packet is dropped,the helicopter is at a horizontal distance $R$ and vertical distance $h$ from the man.The distance $D$ of the helicopter from the man is the hypotenuse of the right-angled triangle formed by $R$ and $h$:$D = \sqrt{R^{2} + h^{2}}$Substituting the value of $R$:$D = \sqrt{\left(v \sqrt{\frac{2h}{g}}\right)^{2} + h^{2}}$$D = \sqrt{\frac{2v^{2}h}{g} + h^{2}}$

$A$ helicopter is flying horizontally with a speed $v$ at an altitude $h$ and has to drop a food packet for a man on the ground. What is the distance of the helicopter from the man when the food packet is dropped?

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