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16 April, 21:32

A projectile is launched from ground level at an angle of 13.0 ° above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?

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  1. 16 April, 22:27
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    The launch angle should be adjusted to 30.63°

    Step-by-step explanation:

    The range of a projectile which is the horizontal distance covered by the projectile can be expressed as;

    R = (v^2 sin2θ) / g

    Where

    R = range

    v = initial speed

    θ = launch angle

    g = acceleration due to gravity

    For the case above. When the projectile is launched at angle 13° above the horizontal.

    θ1 = 13

    R1 = (v^2 sin2θ1) / g

    R1 = (v^2 sin26°) / g ... 1

    For the range to double

    R2 = (v^2 sin2θ) / g ... 2

    R2 = 2R1

    Substituting R2 and R1

    (v^2 sin2θ) / g = 2 * (v^2 sin26°) / g

    Divide both sides by v^2/g

    sin2θ = 2sin26

    2θ = sininverse (2sin26)

    θ = sininverse (2sin26) / 2

    θ = 30.63°
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