? 發問於 科學及數學其他 - 科學 · 10 年 前

Mechanics

A ball is thrown with a speed of 12ms^-1 at an anglr of 30 degree to the horizontal. It is initially at ground level.

Find the speed and direction of flight of the ball after 0.5s and 0.1s

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  • 10 年 前
    最佳解答

    Neglecting air resistance, the horizontal component of the ball's velocity is a constant which is 12 cos 30 = 6√3 m/s

    Then taking g = 10 m/s2 for the vertical part of motion:

    Initial vertical velocity = 12 sin 30 = 6 m/s

    When t = 0.1 s:

    v = 6 - 0.1 x 10 = 5 m/s (upward)

    So magnitude = √[52 + (6√3)2] = 11.53 m/s

    with angle of elevation = tan-1 (5/6√3) = 25.7 deg.

    When t = 0.5 s:

    v = 6 - 0.5 x 10 = 1 m/s (upward)

    So magnitude = √[12 + (6√3)2] = 10.44 m/s

    with angle of elevation = tan-1 (1/6√3) = 5.5 deg.

    資料來源: Myself
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