Question
a projectile is fired with an initial speed of 53 m/s. Find the angle of projection such that the maximum height of the projectile is equal to its horizontal range
Answers
range is v^2 sin(2θ)/g
height is v^2 sin^2(θ)/2g
so,
sin(2θ) = sin^2(θ)/2
2sinθcosθ = sin^2(θ)/2
4cosθ = sinθ
16cos^2(θ) = sin^2(θ) = 1 - cos^2(θ)
17cos^2(θ) = 1
cosθ = 1/√17
height is v^2 sin^2(θ)/2g
so,
sin(2θ) = sin^2(θ)/2
2sinθcosθ = sin^2(θ)/2
4cosθ = sinθ
16cos^2(θ) = sin^2(θ) = 1 - cos^2(θ)
17cos^2(θ) = 1
cosθ = 1/√17
0.34 & 0.125
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