A fighter plane flying horizontally at an alt | vedclass

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(D) Height of the fighter plane $h = 1.5 \; km = 1500 \; m$.Speed of the fighter plane $v = 720 \; km/h = 720 	imes (5/18) \; m/s = 200 \; m/s$.Let $

Vedclass · Accepted Answer

$19.5^o$ and $16 \; km$

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(D) Height of the fighter plane $h = 1.5 \; km = 1500 \; m$.Speed of the fighter plane $v = 720 \; km/h = 720 	imes (5/18) \; m/s = 200 \; m/s$.Let $	heta$ be the angle with the vertical so that the shell hits the plane.Muzzle velocity of the gun $u = 600 \; m/s$.Let $t$ be the time taken by the shell to hit the plane.Horizontal distance travelled by the shell $= (u \sin 	heta) t$.Horizontal distance travelled by the plane $= v t$.For the shell to hit the plane,the horizontal distances must be equal:$(u \sin 	heta) t = v t \implies \sin 	heta = v/u$.$\sin 	heta = 200 / 600 = 1/3 \approx 0.333$.$	heta = \sin^{-1}(0.333) \approx 19.5^o$.To avoid being hit,the pilot must fly at an altitude $H$ higher than the maximum height reached by the shell.The vertical component of the shell's velocity is $u_y = u \cos 	heta = 600 \cos(19.5^o) \approx 600 	imes 0.9426 \approx 565.56 \; m/s$.The maximum height $H$ reached by the shell is $H = u_y^2 / (2g) = (565.56)^2 / (2 	imes 10) = 319858 / 20 \approx 15992.9 \; m \approx 16 \; km$.

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