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Motion in a Plane

Section: Physics

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Question : 60 of 94

Marks: +1, -0

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?

[31 Aug 2021 Shift 1]

$\sqrt{\frac{2 g h v^{2}+1}{h^{2}}}$
$\sqrt{2 g h v^{2}+h^{2}}$
$\sqrt{\frac{2 v^{2} h}{g}+h^{2}}$
$\sqrt{\frac{2 g h}{v^{2}}}+h^{2}$

Solution:

Given, horizontal velocity of helicopter,

u_{x}=v

Height of the helicopter from ground = h

The given condition can be drawn in the figure given below

Here, M is the position of man.

Let, time taken by food packet to reach ground from helicopter = t

Acceleration due to gravity,

g=10\,\mathrm{ms}^{-2}

Initial velocity of food packet along Y-axis,

u_{y}=0

Since,

h=u_{y} t+\frac{1}{2} g t^{2}

\therefore t=\sqrt{\frac{2h}{g}}

and distance travelled by helicopter in time t along horizontal direction

\Rightarrow x=u_{x} t

=v \sqrt{\frac{2h}{g}}

Hence, distance of helicopter from man,

H=\sqrt{h^{2}+x^{2}}=\sqrt{h^{2}+\left(v \sqrt{\frac{2h}{g}}\right)^{2}}

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

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