A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Neglect air resis

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A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Neglect air resistance and use g=9.8m/sec2 as the acceleration of gravity.

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  1. Answer:

    θ₀ = 84.78° (OR) 5.22°

    Explanation:

    This situation can be treated as projectile motion. The parameters of this projectile motion are:

    R = Range of Projectile = 150 m

    V₀ = Launch Speed of Projectile = 90 m/s

    g = 9.8 m/s²

    θ₀ = Launch angle (OR) Angle of Elevation = ?

    The formula for range of a projectile is given as:

    R = V₀² Sin 2θ₀/g

    Sin 2θ₀ = Rg/V₀²

    Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²

    2θ₀ = Sin⁻¹ (0.18)

    θ₀ = 10.45°/2

    θ₀ = 5.22°

    Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:

    θ₀ = 90° – 5.22°

    θ₀ = 84.78°

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