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An aeroplane moving horizontally with a speed of 180km/h drops a food packet while flying at a height of 490m. The horizontal range of the packet is:
(A) 180 m
(B) 980 m
(C) 500 m
(D) 670 m

seo-qna
Last updated date: 13th Jun 2024
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Answer
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Hint:The plane is moving in the horizontal direction with a constant speed and from it, a parcel is dropped. From Newton’s law of inertia, the parcel tends to move in the horizontal direction, so it will not fall straight in the vertical path on to the ground.

Complete step by step answer:
Velocity of plane= initial velocity of parcel= 180 km/h= \[180\times \dfrac{5}{18}=50 m/s\]
The height from the ground, h= 490 m
We need to find the range of the parcel covered.
Splitting the velocity into its components
\[\begin{align}
&\Rightarrow {{u}_{x}}=50 m/s \\
&\Rightarrow {{u}_{y}}=0 m/s \\
\end{align}\]
Considering the vertical motion using Newton’s second equation of motion, there is no acceleration in horizontal direction since the plane was moving with a constant speed and there is constant acceleration in the vertical direction which is the acceleration due to gravity.
\[\begin{align}
&\Rightarrow y={{u}_{y}}t+\dfrac{a{{t}^{2}}}{2} \\
&\Rightarrow 490=0+\dfrac{9.8\times {{t}^{2}}}{2} \\
&\Rightarrow {{t}^{2}}=\dfrac{490}{9.8} \\
&\therefore t=10 s \\
\end{align}\]
So the time taken to reach the ground is 10 s and the horizontal speed was 50 m/s so the horizontal distance covered will be R=ut
\[\begin{align}
&\Rightarrow R=50\times 10 \\
&\therefore R =500 m \\
\end{align}\]

So, the correct option is (C).

Note: In this problem, although the parcel was dropped the body in which it was dropped was moving, so the parcel tends to have an initial velocity in the horizontal direction. Also, the only force which acts on the parcel during its flight to the ground is the gravitational pull due to gravity and that acts in a vertical direction.