Q49P - University Physics with Modern Physics

Question

An airplane is flying with a velocity of 90.0m/s at an angle of $\mathbf{23}\mathbf{.}\mathbf{0}\mathbf{°}$above the horizontal. When the plane is 114m directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment. How far from the dog will the suitcase land? Ignore air resistance.

Step-by-Step Solution

Answer

The suitcase will land 795m far from the dog.

Step 1: Introduction

If a particle is thrown in projectile motion, then the velocity of the particle will contain two components, first one is horizontal and other one is vertical.

 

According to the Newton’s laws of motion,

$s=ut+\frac{1}{2}a{t}^2$ 

Here s is the displacement,u is the initial velocity, a is the acceleration and t is the time.

Step 2: Given data

Velocity = 90.0m/s 

      

Angle = $23.0°$

 

Distance = 114 m    

Step 3: Explanation

For the vertical motion velocity component is $v \sin \theta$ and for the horizontal motion the velocity component is $v \cos \theta$.

      

For the vertical motion from the second law of motion,

$\mathrm{s}=\mathrm{vsin\theta t}+\frac{1}{2}{\mathrm{at}}^2$      

 

Substitute 114m for s, 90m/s for v, $23°$ for $\theta$ and $9.8\mathrm{m}/{\mathrm{s}}^2$ for a in the above equation.

$$\begin{gathered} 114=90\times \mathrm{t}+\frac{1}{2}\times 9.8\times {\mathrm{t}}^2 \\ \mathrm{t}=9.60 \mathrm{s} \end{gathered}$$

 

In the time 9.60s the horizontal distance traveled by suitcase,

$\mathrm{s}=\mathrm{vcos\theta t}+\frac{1}{2}{\mathrm{at}}^2$ 

Substitute 90.0m/s for v, $23°$ for $\theta$, 9.60s for t and $0m/s^2$ for a in the above equation.

$$\begin{gathered} \mathrm{s}=90.0\times \cos 23°\times 9.60+\frac{1}{2}\times 0\times {\left(9.60\right)}^2 \\ \mathrm{s}=795\mathrm{m} \end{gathered}$$       

Therefore the suitcase will land 795m far from the dog.