A projectile’s horizontal range on level ground is R= [(Vo)^2 sin2θ] / g. At what launch angle or angles will the projectile land at half of

Question

A projectile’s horizontal range on level ground is R= [(Vo)^2 sin2θ] / g. At what launch angle or angles will the projectile land at half of its maximum possible range.

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Khánh Gia 2 weeks 2021-08-30T03:35:32+00:00 1 Answers 0 views 0

Answers ( )

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    2021-08-30T03:37:05+00:00

    Answer:

    15° and 75°

    Explanation:

    Given:

    The range of a projectile motion is given as:

    R=\frac{V_0^2sin2\theta}{g}

    Where,

    V_0\to initial\ velocity\\\\\theta\to angle\ of\ projection\\\\g\to\ acceleration\ due\ to\ gravity=9.8\ m/s^2

    Maximum possible range is when \sin 2\theta=1

    So, maximum range possible is, R_{max}=\frac{V_0^2}{g}

    Now, as per question:

    R=\frac{R_{max}}{2}\\\\\frac{V_0^2\sin 2\theta}{g}=\frac{V_0^2}{2g}\\\\\sin 2\theta=\frac{1}{2}\\\\\sin 2\theta=\sin 30\ or\ sin 2\theta=\sin 150\\\\2\theta=30\ or\ 2\theta=150\\\\\theta=\frac{30}{2}=15\ or\ \theta=\frac{150}{2}=75

    Therefore, the launch angles are 15 degree and 75 degree.

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