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4 January, 19:18

A projectile is launched on level ground. At the peak of its trajectory, it is 7.5 m high. It lands

10 m from where it was launched. At what angle was the projectile launched?

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  1. 4 January, 19:30
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    0.5099646386 radians

    Explanation:

    Δy = 7.5 m

    Δx = 10 m

    g = 9.8 m sec⁻²

    From trigonometry,

    Vy₀ = v₀ sin θ

    Vy₀² = v₀² sin²θ

    From the equations of constant acceleration kinematics,

    Vy₀² = 2g Δy

    Two things that are equal to the same thing are equal to each other, so

    v₀² sin²θ = 2g Δy

    v₀² sin²θ = 147 m² sec⁻²

    The double angle formula for the sine,

    sin²θ = 0.5 [1 - cos (2θ) ]

    0.5 v₀² [1 - cos (2θ) ] = 147 m² sec⁻²

    v₀² = 294 m² sec⁻² / [1 - cos (2θ) ]

    The range equation,

    Δx = v₀² sin (2θ) / (2g)

    10 m = 294 m² sec⁻² sin (2θ) / { (2g) [1 - cos (2θ) ]}

    1.5 sin (2θ) = 1 - cos (2θ)

    1.5 sin (2θ) = 1 - √[1-sin² (2θ) ]

    A convenient substitution.

    u = sin (2θ)

    2.25 u² = 1 - 2√[1-u²] + 1 - u²

    3.25 u² - 2 = - 2√[1-u²]

    1 - 1.625u² = √[1-u²]

    1 - 3.25u² + 2.640625u⁴ = 1-u²

    2.640625u⁴ - 2.25u² = 0

    As long as u≠0,

    2.640625u² - 2.25 = 0

    2.640625u² = 2.25

    u = 0.8520710059

    θ = 0.5 arcsin u

    θ = 0.5099646386 radians

    Furthermore,

    v₀ = 15.16666667 m/s
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