Question

If an object is shot with an initial velocity of 160 ft/sec, what is the smallest initial angle will you need to obtain a horizontal distance of 500 feet? (Round to the nearest hundredth of a radian.) Please show work

Answer 1

To solve the question we shall use the formula for the range given by:
Horizontal range, R=[v²sin 2θ]/g
plugging in our values we get:
500=[160²×sin 2θ]/10
5000=160²×sin 2θ
0.1953=sin 2θ
thus:
arcsin 0.1953=2θ
11.263=2θ
hence:
θ=5.6315°~5.63