Question

A food packet is dropped from an aeroplane, moving at a speed of $360km{{h}^{-1}}$ in a horizontal direction, from a height of $500m$. Find (i) its time of descent (ii) the horizontal distance between the point at which the food packet reaches the ground and the point above which it was dropped.

Answer 1

Hint: The motion of the packet dropped from an airplane can be assumed as a vertical projectile motion. The packet gains horizontal velocity equal to the velocity of the airplane and the initial vertical velocity of the packet is zero. The vertical acceleration of the packet is equal to the acceleration due to gravity while the horizontal acceleration is zero. We will apply the equations of motion on the packet’s motion to calculate the time of descent and horizontal distance covered by the packet.

Complete step by step answer:
Consider an airplane moving with a constant speed at an elevated height above the Earth's surface. In the course of its flight on a constant height, the plane drops a packet down from one of its compartments.

The dropped packet follows a parabolic path and remains directly below the plane at all the times. As the packet falls, it undergoes a vertical acceleration during its motion; that is, there is a change in its vertical velocity. This vertical acceleration is due to the downward force of gravity which acts upon the packet. If the packet's motion could be assumed as a projectile motion; that is, if the influence of air resistance could be ignored or assumed negligible, then there would be no horizontal acceleration. In the absence of horizontal forces, the packet would have a constant velocity in the horizontal direction. This explains why the packet is always located directly under the airplane from which it is dropped.
Applying the equation of motion in the vertical direction of the dropped packet,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
As the airplane is in a horizontal motion, the velocity acquired by the packet in vertical direction is zero. The distance covered by the packet is equal to the height at which the plane is moving. The acceleration of the packet in vertical direction is equal to the acceleration due to gravity.
Putting,
$\begin{align}
  & s=h \\
 & u=0 \\
\end{align}$
We get,
$\begin{align}
  & h=0+\dfrac{1}{2}g{{t}^{2}} \\
 & 2h=g{{t}^{2}} \\
 & t=\sqrt{\dfrac{2h}{g}} \\
\end{align}$
Applying the equation of motion in the horizontal direction of the dropped packet,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
As the packet is dropped from a moving airplane, it acquires the horizontal velocity with which the plane is moving. As there is no force acting on the particle in horizontal direction, the acceleration of the particle in horizontal direction is zero. Time taken to travel a horizontal distance is equal to the time of descent.
Putting,
$\begin{align}
  & a=0 \\
 & t=\sqrt{\dfrac{2h}{g}} \\
\end{align}$
We get,
$s=u\sqrt{\dfrac{2h}{g}}$
We are given that a food packet is dropped from an aeroplane, moving a speed of $360km{{h}^{-1}}$ in a horizontal direction, from a height of $500m$,
(i) For calculating time of descent,
 $t=\sqrt{\dfrac{2h}{g}}$
Here,
$h=500m$
Therefore,
$\begin{align}
  & t=\sqrt{\dfrac{2\times 500}{10}}=\sqrt{100} \\
 & t=10s \\
\end{align}$
The time of descent of the packet is $10m{{s}^{-1}}$
(ii) For calculating the horizontal distance travelled by the packet,
 $s=u\sqrt{\dfrac{2h}{g}}$
Here,
$\begin{align}
  & u=360km{{h}^{-1}}=100m{{s}^{-1}} \\
 & t=\sqrt{\dfrac{2h}{g}}=10s \\
\end{align}$
Therefore,
$s=100\times 10=1000m$
The horizontal distance between the point at which the food packet reaches the ground and the point above which it was dropped is $1000m$

Note:
Students should not think that there is horizontal force acting upon the packet since it has a horizontal motion. This is simply not the case. The horizontal motion of the packet is the result of its own inertia. When dropped from the airplane, the packet already possessed a horizontal motion. The packet will maintain this state of horizontal motion unless acted upon by a horizontal force. An object in motion continues to be in motion with the same speed and in the same direction unless acted upon by some force, Newton's first law. Always remember that forces do not cause motion; rather, forces cause accelerations.