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

Question #226544

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?

1
2021-08-23T04:48:55-0400

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

We know

"x = v_ot cos \\theta \\\\\ny = v_ot sin \\theta \u201416t^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 \u201416t) \\\\\nv_osin \\theta -16t = 0 \\\\\nsin 2 \\theta = frac{1}{15}\\\\\n\\frac{1}{2}sin^{-1} (\\frac{1}{15}) =1.91\u00b0 \\\\\n\\implies 1.91\u00b0 \u2264 \\theta \u2264 (90 -1.91)\u00b0"

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