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25 June, 06:49

A projectile is launched over a horizontal surface in such a manner that its maximum height is 3/5 of its horizontal range. Determine the launch angle.

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  1. 25 June, 07:10
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    The launch angle = 67.38°

    Explanation:

    For projectile motion, the range (R) and maximum height (H) reached are represented by the formulas

    R = (u² sin 2θ) / g

    H = (u² sin² θ) / 2g

    H = (3/5) R

    (H/R) = (3/5)

    Dividing the formula for maximum height by that of the range of a projectile

    (H/R) = [ (sin² θ) / 2] : sin 2θ

    Sin 2θ = 2 sinθcosθ

    (H/R) = (sin² θ) / (2*2sinθcosθ)

    (H/R) = (sin² θ) / (4sinθcosθ)

    (H/R) = (sinθ) / (4cosθ) = (3/5)

    Tanθ = (12/5)

    θ = tan⁻¹ (2.4)

    θ = 67.38°
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