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

The horizontal acceleration is = 0.6g
The time taken by the rocket to travel a vertical distance of v=800m is calculated as follows :
Initial vertical velocity is u=0
The acceleration is = g
Apply the equation of motion
\[s=ut+1/2at^2 = 0*t+0.5*9.81*t^2\]
\[800=0.5*9.81*t^2\]
\[t=\sqrt{800/(0.5*9.81)}=12.77s\]
The horizontal travelled in 12.77s is calculated as follows :
Initial velocity is \ \[h=ut+1/2at^2=(1000/3)*12.77 +0.5*0.6*9.81*12.77^2=4736.5m\] The angle is \[\tan \theta= v/h=2400/4736.5=0.507\] The angle is =26.87º