Answer to Question #230422 in Physics for Yzang

Question

A gun has fired at a moving target. The bullet has a muzzle velocity of 1200
ft/s. The gun and the target are both on level ground, with target moving 60 mph
away from the gun have a distance of 30,000 ft from the gun the moment the
bullet was fired. What is the angle of the projectile needed to hit the target?

Answer 1

Given:

$v_{0x}=v_0\cos\theta=1200\cos\theta \;{\rm ft/s}\\ v_{0y}=v_0\sin\theta=1200\sin\theta \;{\rm ft/s}\\ u=60\:{\rm mph}=88{\:\rm ft/s}\\ L=30\:000\:\rm ft$

In the reference frame with respect to the movement target, the range of a bullet is given by

$$L=(v_{0x}-u)t$$

The time of motion

$$t=\frac{2v_{0y}}{g}$$

So, we get

$$L=(1200\cos\theta-88)*\frac{2*1200\sin\theta}{32}=30000$$

$$(1200\cos\theta-88)\sin\theta=400$$

The roots of this equation

$$\cos\theta_1=0.919,\quad \cos\theta_2=0.446$$

Finally

$$\theta_1=23^{\circ},\quad\theta_2=64^{\circ}$$