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?

Expert's answer

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 getÂ

"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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