A projectile is launched over a horizontal surface in such a manner that its maximum height is 1/4 of its horizontal range. Determine the la

Question

A projectile is launched over a horizontal surface in such a manner that its maximum height is 1/4 of its horizontal range. Determine the launch angle.

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Cherry 1 week 2021-09-05T07:11:28+00:00 1 Answers 0 views 0

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    2021-09-05T07:12:36+00:00

    Answer:

    The launch angle \theta=45^o.

    Explanation:

    We know,

    Horizontal distance travelled in projectile motion , R=\dfrac{u^2sin2\theta}{g}.

    Also, Maximum height ,H_{max}=\dfrac{u^2sin^2\theta}{2g}.

    Now, according to question maximum height is 1/4 of its horizontal range.

    H_{max}=\dfrac{1}{4}\times R

    \dfrac{u^2 sin^2\theta}{2g}=\dfrac{1}{4}\times \dfrac{u^2 sin\ 2\theta}{g}

    Simplifying above expression.

    We get , \theta=45^o

    Hence, this is the required solution.

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