A physics Q - Please Help!!~ (with explaination) Thank~

A water rocket moves vertically upward from rest with an acceleration of 15 ms-2 when water is being ejected. The ejecting of water lasts for 2 s. It then moves freely under gravity. What is the maximum heigth reached by the water rocket?

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  • 1 十年前
    最愛解答

    Take upward direction be positive.

    Then, find out the height gained by the rocket in the first 2 seconds

    By s = ut + 1/2 at2

    s = (0)(2) + 1/2 (15)(2)2

    s = 30 m

    Therefore, the height gained by the rocket in the first 2 seconds is 30 m.

    By v = u + at

    v = (0) + 15(2)

    v = 30 ms-1

    So, the velocity of the rockect is 30 ms-1 upward.

    After 2 seconds, the rocket experiences a downward force due to its weight. That is, it will acclerate downward by free falling.

    At maximum height, the instantaneous velocity of the rocket is 0.

    So, by v2 = u2 + 2as

    (0)2 = (30)2 + 2(-10)s

    s = 45 m

    So, after the ejecting of water, the rocket can still go up by 45 m.

    As a result, maximum height gained by the water rocket

    = 30 + 45

    = 75 m

    資料來源: Myself~~~
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