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working with projectile motion- finding angle of projection

A projectile is fired in such a way that its horizontal range is equal to three times its maximum height. What is the angle of projection?

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Please see the attachment to see how you can resolve velocity vector in to components.
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<br>The horizontal range is determined by the horizontal component of the velocity which is Vx and the maximum height is determined by the y component of the velocity Vy
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<br>Vx = V Cos(A) where A is the angle of projection
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<br>Vy = V Sin(A)
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<br>Vx is a constant but Vy is continuously getting changed as the body flies.
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<br>The horizontal range is given by the expression,
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<br> R = velocity * time = Vx * t ......(1)
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<br>Where t is the total time of flight. Since the object has to go up and down, its total displacement is 2h but its average velocity during this total displacement is 1/2*(Vy).
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<br>So the travel time t is
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<br>t = (2h)/(1/2)(Vy) = 4h/Vy ....(2)
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<br>But from (1), t = R/Vx
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<br>Equate both the equations
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<br>R/Vx = 4h/Vy
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<br>But it is given that R = 3h
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<br>3h/Vx = 4 h /Vy
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<br>Or, Vy/Vx = 4/3
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<br>But , Vx = V Cos A and Vy = V Sin(A)
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<br>thus we get, SinA/CosA = 4/3
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<br>or, tan A = 4/3
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<br>from this A = tan^-1 (4/3) = 53.13 Degree

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