Question

An airplane flying horizontally at a constant speed of 350 km/h over level ground releases a bundle of food supplies. Ignore the effect of the air on the bundle. What are the bundle’s 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, 450 km/h , would the time of fall be longer, shorter, or the same?

Short answer

1. The vertical component of the velocity is $v_{0y}=0$.
2. The Horizontal component of the velocity is $v_{0x}=350\mathrm{km}/\mathrm{h}$.
3. The Horizontal component of velocity just before hitting the ground is $v_{fx}=350\mathrm{km}/\mathrm{h}$.
4. The time of fall will be same.

Step-by-step solution

Given information

Airplane speed is 350 km/h ( horizontally) .

To understand the concept

It deals with the kinematic equations of motion in which the motion can be described with constant acceleration. Also, this problem is based on the concept of projectile motion. When a particle is thrown near the earth’s surface, it travels along a curved path under constant acceleration directed towards the center of the earth surface. This is a projectile motion problem, which contains two independent motions.

Formulae:

Kinematic equations:

$$\begin{gathered} v_{fy}=v_{0y}+at \\ ∆y=y_{0y}t+\left(\frac{1}{2}{a}_y{t}^2\right) \end{gathered}$$

(a) To find Vertical component of initial velocity.

Airplane is moving horizontally; hence initially bundle has speed only along horizontal direction.

Hence,

$v_{0y}=0$

(b) To find horizontal component of initial velocity

And horizontal component will be same as airplanes speed as it was in the plane moving along with it.

Hence,

$v_{0x}=350\mathrm{km}/\mathrm{h}$

(c) To find horizontal component of velocity just before hitting the ground

It is known that, horizontal motion would occur with constant speed as there is no acceleration in the horizontal direction.

Hence,

$$\begin{gathered} v_{fx}=v_{0x} \\ =350\mathrm{km}/\mathrm{h} \end{gathered}$$

(d) To check the time of fall, when speed is

$$\begin{gathered} ∆y=v_{0y}t+\left(1/2a_y{t}^2\right) \\ \mathrm{As},V_{iy}=0 \\ ∆y=\left(\frac{1}{2}{a}_y{t}^2\right) \end{gathered}$$

From this formula we can say that, time of fall depends on vertical height only, hence it would be same for any speed.