Q74P - Fundamentals Of Physics

Question

A pilot flies horizontally at $\mathbf{1300}\mathbf{km}\mathbf{/}\mathbf{h}$, at height $\mathbf{h}\mathbf{=}\mathbf{35}\mathbf{ }\mathbf{m}$above initially level ground. However, at time t=0 , the pilot begins to fly over ground sloping upward at angle $\mathbf{\theta }\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{3}\mathbf{°}$(Fig. 2-41). If the pilot does not change the airplane’s heading, at what time   does the plane strike the ground?

Step-by-Step Solution

Answer

The time t at which the plane strikes the ground is 1.29 s

Step 1: Given data

 The speed of the plane is $\mathrm{v}=1300 \mathrm{km}/\mathrm{h}$ at height $\mathrm{h}=35 \mathrm{m}$above the ground.

The angle sloping upward is $\theta =4.3°$.

Step 2: Relation between speed, distance, and time at constant acceleration

When acceleration is constant one can say that speed is the ratio of distance and time.

Step 3: Calculation of distance using trigonometry

If the plane moves with velocity $\mathrm{v}=1300\mathrm{km}/\mathrm{h}$ continuously and the terrain continues its upward slope of $4.3°$, then the plane will strike the ground after traveling a distance of

$$\begin{gathered} \mathrm{x}=\frac{\mathrm{h}}{\tan 4.3} \\ =\frac{35\mathrm{m}}{\tan 4.3} \\ =465.5 \mathrm{m} \end{gathered}$$ 

 So, the time of flight is,

 $$\begin{gathered} \mathrm{t}=\frac{\mathrm{x}}{\mathrm{v}} \\ =\frac{0.465\mathrm{km}}{1300\mathrm{km}/\mathrm{h}} \\ =0.00036\mathrm{hr} \\ =1.29 \mathrm{s} \\ \approx 1.3 s \end{gathered}$$

 Therefore, at 1.3 s the plane will hit the ground.