Asked by eric

A projectile is fired in such a way that its horizontal range is equal to 8.5 times its maximum height. What is the angle of projection?
Answered by tchrwill
From h = V^2(sin2µ)/2g and d = V^2(sin^2µ)/g:

d = V^2(sin2µ)/32 = 8.5V^2(sin^2µ)/32 yielding µ = 25.2º.
Answered by Anonymous
A projectile is fired in such a way that its horizontal range is equal to 2.5 times its maximum height. What is the angle of projection?
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