A fighter plane enters inside the enemy territory, at time $t = 0$ with velocity $v_{0} = 250\ ms^{- 1}$ and moves horizontally with constant acceleration $a = 20ms^{- 2}$ (see figure). An enemy tank at the border, spot the plane and fire shots at an angle $ heta = 60{^\circ}$ with the horizontal and with velocity $u = 600\ ms^{- 1}$. At what altitude $H$ of the plane it can be hit by the shot? [image:4_image4.emf]

Question

A fighter plane enters inside the enemy territory, at time $t = 0$ with velocity $v_{0} = 250\ ms^{- 1}$ and moves horizontally with constant acceleration $a = 20ms^{- 2}$ (see figure). An enemy tank at the border, spot the plane and fire shots at an angle $	heta = 60{^\circ}$ with the horizontal and with velocity $u = 600\ ms^{- 1}$. At what altitude $H$ of the plane it can be hit by the shot? [image:4_image4.emf]

Dr. Abubacker · Accepted Answer

(a) If it is being hit then
$d = v_{0}t + \frac{1}{2}at^{2} = (u\cos	heta)t$
or $t = \frac{u\cos	heta - v_{0}}{a/2}$
[image:4_image89.emf]
$	herefore\ \ t = \frac{600 	imes \frac{1}{2} - 250}{10} = 5\ s$
$H = \left( u\sin	heta ight)t - \frac{1}{2} 	imes gt^{2}$
$= 600 	imes \frac{\sqrt{3}}{2} 	imes 5 - \frac{1}{2} 	imes 10 	imes 25$
$H = 2473\ m$

Explanation

Reviewed by Dr. Abubacker

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