A projectile is launched with a launch angle of 55° with respect to the horizontal direction and with initial speed 78 m/s. How long does it

Question

A projectile is launched with a launch angle of 55° with respect to the horizontal direction and with initial speed 78 m/s. How long does it remain in flight?

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Linh Đan 5 years 2021-09-04T15:31:51+00:00 1 Answers 20 views 0

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  1. ngockhue
    0
    2021-09-04T15:33:41+00:00
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    Answer:

    The projectile is in air for 13.03 seconds.

    Explanation:

    Given that,

    Angle of projection of the projectile, \theta=55^{\circ}

    Initial speed of the projectile, u = 78 m/s

    To find,

    We need to find the time of flight of the projectile.

    Solution,

    It is defined as the time taken by the projectile when it is in air. It is given by the formula as :

    T=\dfrac{2u\ \sin\theta}{g}

    T=\dfrac{2\times 78\ \sin(55)}{9.8}

    T = 13.03 seconds

    So, the projectile is in air for 13.03 seconds.

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