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 $g=10m{s}^{-2}$)

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Solution

## Step 1: Given data Height at which plane is flying, $h=1.5km=1500m$Speed of plane, $v=720km{h}^{-1}=720×\frac{5}{18}=200m{s}^{-1}$Speed of bullet of firing gun, $u=600m{s}^{-1}$Acceleration due to gravity, $a=10m{s}^{-2}$Step 2: CalculationStep 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 $vt={u}_{x}t\phantom{\rule{0ex}{0ex}}⇒v={u}_{x}\phantom{\rule{0ex}{0ex}}⇒v=u\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\frac{v}{u}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\frac{200}{600}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =0.33\phantom{\rule{0ex}{0ex}}⇒\theta ={\mathrm{sin}}^{-1}\left(0.33\right)\phantom{\rule{0ex}{0ex}}⇒\theta =19.{27}^{\circ }$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 .${H}_{max}=\frac{{u}^{2}{\mathrm{sin}}^{2}\left({90}^{o}-\theta \right)}{2g}=\frac{600×600×0.89}{2×10}=16.02km$ 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 $16.02km$.

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