Question

A fighter plane flying horizontally at an altitude of $1.5km$ with a speed of $720km{h}^{-1}$ passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired the shell with muzzle speed $600m{s}^{-1}$ to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take $g=10m{s}^{-2}$)

Answer 1

Step 1: Given data

Height at which plane is flying, $h=1.5km=1500m$

Speed of plane, $v=720km{h}^{-1}=720\times \frac{5}{18}=200m{s}^{-1}$

Speed of bullet of firing gun, $u=600m{s}^{-1}$

Acceleration due to gravity, $a=10m{s}^{-2}$

Step 2: Calculation

Step 3: Finding the angle at which a gun should be fired for the plane to hit its target, $\theta$

Distance traveled by plane =$vt$ ($t=$time taken by the shell to hit the plane )

Distance traveled by the shell in x-direction = $u_{x}t$

These $2$ distances are equal since the shell hits the plane

$$\begin{gathered} vt=u_{x}t \\ \Rightarrow v=u_x \\ \Rightarrow v=u\sin \theta \\ \Rightarrow \sin \theta =\frac{v}{u} \\ \Rightarrow \sin \theta =\frac{200}{600} \\ \Rightarrow \sin \theta =0.33 \\ \Rightarrow \theta ={\sin }^{-1}(0.33) \\ \Rightarrow \theta =19.{27}^{\circ } \end{gathered}$$

Step 4: Finding the height at which the pilot should fly the plane to avoid getting hit.

The pilot should fly the plane at a height more than the maximum height of launch of the shell at any angle

According to projectile motion .$H_{max}=\frac{u^2{\sin }^2({90}^o-\theta )}{2g}=\frac{600\times 600\times 0.89}{2\times 10}=16.02km$

The angle at which the shell should be launched is $19.{27}^{\circ }$and the height at which the plane should be kept to avoid the shell attack should be more than $16.02km$.