Asked by junoo

find the angle of projection at the horizontal range is twice the maximum height of a projecyile
Answered by Chanz
Range is x = vo^2 sin2θ/g
Max Height is y = (vo Sinθ)^2/2g
Set range = twice max height
The vo^2 will cancel, solve for θ
Answered by Anonymous
find the angle of projection at which the horizontal range is twice the maximum teight of a projective.
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