Q54P - University Physics with Modern Physics

Question

An airplane is dropping bales of hay to cattle stranded in a blizzard on the Great Plains. The pilot releases the bales at 150 m above the level ground when the plane is flying at 75 m/s in a direction $\mathbf{55}\mathbf{°}$above the horizontal. How far in front of the cattle should the pilot release the hay so that the bales land at the point where the cattle are stranded?

Step-by-Step Solution

Answer

For the landing of bales land at the point where the cattle have been stranded the distance should be 395.79 m far in front of the cattle.

Step 1: A concept of equation of motion:

The velocity contains two components, one is a horizontal component and another one is vertical. 

 

According to the newton’s laws of motion,

$\mathbf{s}\mathbf{=}\mathbf{ut}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{at}}^{\mathbf{2}}$ 

 Where, s , u , t , and a are displacement, initial velocity, time, and acceleration respectively.

Step 2: As given data:

Vertical distance, ${\mathrm{s}}_{\mathrm{v}}=150 \mathrm{m}$ 

 

Angle, $\mathrm{\theta }=55°$  

 

Initial velocity, u=75 m/s          

Step 3: Define the required distance:

For the vertical motion, gravity is negative. Therefore, solve equation of motion for time as below.

      

$$\begin{gathered} {\mathrm{s}}_{\mathrm{v}}=\mathrm{usin\theta t}-\frac{1}{2}{\mathrm{gt}}^2 \\ -150 \mathrm{m}=\left(75 \mathrm{m}/\mathrm{s}\times \sin 55°\times \mathrm{t}\right)-\left(\frac{1}{2}\times 9.8 \mathrm{m}/{\mathrm{s}}^2\times {\mathrm{t}}^2\right) \\ -150 \mathrm{m}=\left(61.425\times \mathrm{t}\right)-\left(4.9\times {\mathrm{t}}^2\right) \\ 4.9\times {\mathrm{t}}^2-61.425\times \mathrm{t}+150=0 \end{gathered}$$ 

 

By using quadratic solution, you obtain

$$\begin{gathered} \mathrm{t}=\frac{-\left(-61.425\right)\pm \sqrt{{\left(-61.425\right)}^2-4\times 4.9\times 150}}{2\times 4.9} \\ =\frac{-\left(-61.425\right)\pm \sqrt{3773.031-2940}}{9.8} \\ =\frac{61.425\pm 28.862}{9.8} \\ =6,27\pm 2.94 \mathrm{s} \end{gathered}$$ 

 

Take a positive sign, and you have

$$\begin{gathered} \mathrm{t}=(6.27 +2.94) \mathrm{s} \\ =9.21 \mathrm{s} \end{gathered}$$ 

 

For the horizontal motion, gravity is zero

$$\begin{gathered} {\mathrm{s}}_{\mathrm{H}}=\mathrm{ucos\theta t} \\ =75 \mathrm{m}/\mathrm{s}\times \cos 55°\times 9.21\mathrm{s} \\ =395.79 \mathrm{m} \end{gathered}$$                                                                                     

Hence, for the landing of bales land at the point where the cattle have stranded the distance should be 395.79 m far in front of the cattle.