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A fighter plane flying horizontally at an altitude of 1.5 $\mathrm{km}$ with speed $720 \mathrm{~km} / \mathrm{h}$ passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed $600 \mathrm{~ms}^{-1}$ to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take $g=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
Velocity of plane,
$$
v_p=720 \times \frac{5}{18} \mathrm{~ms}^{-1}=200 \mathrm{~ms}^{-1}
$$
Velocity of shell $=600 \mathrm{~ms}^{-1}$;
$\sin \theta=\frac{200}{600}=\frac{1}{3}$
or $\theta=\sin ^{-1}\left(\frac{1}{3}\right)=19.47^{\circ}$
This angle is with the vertical.
Let $h$ be the required minimum height. Using equation $v^2-u^2=2 a s$, we get
$(0)^2-(600 \cos \theta)^2=-2 \times 10 \times h$
or, $h=\frac{600 \times 600\left(1-\sin ^2 \theta\right)}{20}$
$$
=30 \times 600\left(1-\frac{1}{9}\right)=\frac{8}{9} \times 30 \times 600 \mathrm{~m}=16 \mathrm{~km} \text {. }
$$
$$
v_p=720 \times \frac{5}{18} \mathrm{~ms}^{-1}=200 \mathrm{~ms}^{-1}
$$
Velocity of shell $=600 \mathrm{~ms}^{-1}$;
$\sin \theta=\frac{200}{600}=\frac{1}{3}$
or $\theta=\sin ^{-1}\left(\frac{1}{3}\right)=19.47^{\circ}$
This angle is with the vertical.
Let $h$ be the required minimum height. Using equation $v^2-u^2=2 a s$, we get
$(0)^2-(600 \cos \theta)^2=-2 \times 10 \times h$
or, $h=\frac{600 \times 600\left(1-\sin ^2 \theta\right)}{20}$
$$
=30 \times 600\left(1-\frac{1}{9}\right)=\frac{8}{9} \times 30 \times 600 \mathrm{~m}=16 \mathrm{~km} \text {. }
$$
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