JEE MAIN - Physics (2021 - 31st August Morning Shift - No. 1)
A helicopter is flying horizontally with a speed 'v' at an altitude 'h' has to drop a food packet for a man on the ground. What is the distance of helicopter from the man when the food packet is dropped?
$$\sqrt {{{2gh{v^2} + 1} \over {{h^2}}}} $$
$$\sqrt {2gh{v^2} + {h^2}} $$
$$\sqrt {{{2{v^2}h} \over g} + {h^2}} $$
$$\sqrt {{{2gh} \over {{v^2}}} + {h^2}} $$
Explanation
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$$R = \sqrt {{{2h} \over g}} .\,v$$
$$D = \sqrt {{R^2} + {h^2}} $$
$$ = \sqrt {{{\left( {\sqrt {{{2h} \over g}} .\,v} \right)}^2} + {h^2}} $$
$$D = \sqrt {{{2h{v^2}} \over g} + {h^2}} $$
Option (c) is correct.
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