Answer to Question #226544 in Civil and Environmental Engineering for camille

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

We are given $(x,y)= (30,000)$ and; $v_o =1200ft /s$ .

We know

$x = v_ot cos \theta \\ y = v_ot sin \theta —16t^2$

From equation (1) (by putting values), we get 

$t=\frac{3000 } {1200 cos \theta}$

Using this value on second equation, we get;

$0 = t(v_o sin \theta —16t) \\ v_osin \theta -16t = 0 \\ sin 2 \theta = frac{1}{15}\\ \frac{1}{2}sin^{-1} (\frac{1}{15}) =1.91° \\ \implies 1.91° ≤ \theta ≤ (90 -1.91)°$