Question

A rocket is released a jet fighter flying horizontally at 1200 kph at an altitude of 2400m above its target. The rocket thrust gives it a constant acceleration of 0.6g. Determine the angle between the horizontal and the line of the sight to the target.

Answer 1

Initial velocity of rocket is = u=1200*1000/3600= 333.3m/s

Horizontal acceleration of rocket is a=0.6 m/s^2

Horizontal distance to the target is

d=333.3*t+0.5*0.6g*t^2

Vertical acceleration of rocket is = g m/s^2

Initial vertical velocity of rocket is u =0m/s

Therefore,

2400=0*t+0.5*g*t^2

t^2=2400/(0.5*g)=489.8s^2

t=sqrt(489.8)=22.13s

Therefore,

d=333.3*22.13+0.5*0.6*g*22.13^2 = 8816.4 m

The angle is

= arctan(2400/8816.4)=15.2º