Question

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

Answer 1

Step 1: Draw rough sketch of given situation.

$h=1.5\text{km}=1500m,v=720\text{km/h}=200m/s$.

Step 2: Find angle of projection with vertical.
Muzzle velocity of the gun, $u=600m/s$
Horizontal component of velocity, $u_x=u\sin \theta$
So, Horizontal distance travelled by the shell $=u_{x}t\dots \left(1\right)$
Distance travelled by the plane during time t = vt
Shell hits the plane if these two distances are equal.
$u_{x}t=vt$
$\Rightarrow \sin \theta =0.33$
$\theta ={\sin }^{-1}\frac{1}{3}$

Step 3: Find maximum height up to which the shell can hit the plane.
Maximum height up to which the shell can hit the plane will be
$y_{max}=\frac{u^2{\sin }^2({90}^{\circ }-\theta )}{2g}$
$y_{max}=\frac{u^2{\cos }^2\theta }{2g}$
$y_{max}=\frac{\left(600{)}^2\right\{1-\left(0.33{)}^2\right\}}{2\times 10}$
$y_{max}=16040m\approx 16km$.

Hence, the pilot should fly the plane at a minimum height of 16 km to avoid being hit.