Question

A fighter plane flying horizontally at an altitude of $1.5\mathrm{km}$ with a speed of $200{\mathrm{ms}}^{-1}$ passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell to hit the plane. The speed of the shell is $600{\mathrm{ms}}^{-1}$. At what minimum altitude should the pilot fly the plane to avoid being hit ? [Take $g=10{\mathrm{ms}}^{-2}$ ]

Answer 1

Step 1: Given data

The altitude of the fighter plane is $1.5\mathrm{km}=1500\text{m}$

Speed of the plane, $v=200{\text{ms}}^{-1}$

The speed of the shell is $u=600{\mathrm{ms}}^{-1}$

Step 2: Find the angle from the vertical should the gun be fired for the shell to hit the plane.

Consider the time taken for the shell to hit the plane is $t$.

The distance travelled by the plane is ${\mathit{v}}_{\mathbf{x}}\mathbf{\times }\mathit{t}\mathbf{=}\mathit{v}\mathit{t}$ because it is flying horizontally.

The horizontal distance travelled by the gun fired from the shell is ${\mathit{u}}_{\mathbf{x}}\mathit{t}\mathbf{=}\mathit{u}\mathbf{}\mathbf{cos}\mathit{\theta }\mathbf{\times }\mathit{t}$ .

Here, $u_x$ is the horizontal component of the initial velocity of the gun fired and $\theta$ is the angle from the vertical at which it is fired.

For the shell to hit the plane, the distance travelled by plane and the shell should be equal.

$$\begin{gathered} \therefore u\cos \theta \times t=vt \\ \Rightarrow u\cos \theta =v \\ \Rightarrow 600\times \cos \theta =200 \\ \Rightarrow cos\theta =\frac{1}{3} \\ \Rightarrow \theta =co{s}^{-1}\left(\frac{1}{3}\right) \\ \Rightarrow \theta =70.53° \end{gathered}$$

Step 3: Find the maximum height of the shell

To avoid being hit by the shell, the plane should fly higher than the maximum height of the shell.

The maximum height of the shell is given as $\mathit{H}\mathbf{=}\frac{{\mathbf{u}}^{\mathbf{2}}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{(}\mathbf{90}\mathbf{°}\mathbf{-}\mathbf{\theta }\mathbf{)}}{\mathbf{2}\mathbf{g}}$. Here, $\theta$ is the angle from the vertical

We had taken the value of the angle as $\mathbf{(}\mathbf{90}\mathbf{°}\mathbf{-}\mathbf{\theta }\mathbf{)}$ because the maximum height formula uses the value of the angle from the horizontal but $\theta$ is the angle from the vertical.

$$\begin{gathered} \therefore H=\frac{{600}^2{\sin }^2(90°-70.53°)}{20} \\ =\frac{{600}^2{\sin }^2(70.53°)}{20} \\ =16876.14\text{m} \\ =1.69\times {10}^4\text{m} \end{gathered}$$

Hence, at an angle of ${\left[co{s}^{-1}\left(\frac{1}{3}\right)\right]}^{\mathbf{°}}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathbf{70}\mathbf{.}\mathbf{53}\mathbf{°}$ from the vertical, the gun should be fired for the shell to hit the plane. Also, $\mathbf{1}\mathbf{.}\mathbf{69}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathit{m}$ is the minimum altitude at which the pilot should fly the plane to avoid being hit by the shell.