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# An airplane flying horizontally at a constant speed of ${\text{350}}{{{\text{km}}} \mathord{\left/ {\vphantom {{{\text{km}}} {\text{h}}}} \right.} {\text{h}}}$ over level ground releases a bundle of food supplies. Ignore the effect of the air on the bundle. What are the bundles initial (a) vertical and (b) horizontal components of velocity? (c) What is its horizontal component of velocity just before hitting the ground? (d) If the airplane's speed were, instead, ${\text{450}}{{{\text{km}}} \mathord{\left/ {\vphantom {{{\text{km}}} {\text{h}}}} \right.} {\text{h}}}$ , would the time of fall be longer, shorter, or the same?

Last updated date: 14th Sep 2024
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Hint: There is no significant horizontal acceleration before the bundle touches the ground. Time of fall is determined by vertical acceleration and vertical initial velocity.

Complete Step by Step Solution
It has been given that, an airplane flying horizontally at a constant speed of ${\text{350}}{{{\text{km}}} \mathord{\left/ {\vphantom {{{\text{km}}} {\text{h}}}} \right. } {\text{h}}}$ over level ground releases a bundle of food supplies.
The initial velocity of food bundle is zero because the plane is moving horizontally and its vertical component is zero.
Thus, initial velocity $u = 0{{km} \mathord{\left/ {\vphantom {{km} h}} \right. } h}$
Since the food bundle is inside the aeroplane, which is moving horizontally with a constant speed, the velocity of the food bundle is the same as that of the aeroplane.
The food bundle's initial and uniform horizontal velocity is ${\text{350}}{{{\text{km}}} \mathord{\left/ {\vphantom {{{\text{km}}} {\text{h}}}} \right. } {\text{h}}}$ .
The food bundle will maintain its horizontal velocity all along its journey until it hits the ground. The reason behind this is that the bundle does not face any force which acts opposite to it.
The only force that increases the velocity of the bundle vertically and not horizontally, is the vertical gravitational force. So the horizontal component of its velocity just before hitting the ground is ${\text{350}}{{{\text{km}}} \mathord{\left/ {\vphantom {{{\text{km}}} {\text{h}}}} \right. } {\text{h}}}$ .
If the horizontal velocity of plane is increased to ${\text{450}}{{{\text{km}}} \mathord{\left/ {\vphantom {{{\text{km}}} {\text{h}}}} \right. } {\text{h}}}$ . There will be no effect on time of fall of the bundle. This is because the horizontal velocity has nothing to do with vertical displacement. In fact, the vertical displacement is due to gravitational acceleration $g$ . So the time of fall will be the same.

Note
For a projectile motion, displacement can be calculated from the point of projection along the direction of projection. Velocity $v$ has two components $v\cos \theta$ in horizontal and $v\sin \theta$ in vertical direction, in which horizontal component of velocity remains constant because of absence of force. Acceleration due to gravity is present throughout the motion in vertical direction.