A projectile is launched from the top of a tower of height 30m with an initial velocity of 20m/s at an angle of 30degree above the horizonta

Question

A projectile is launched from the top of a tower of height 30m with an initial velocity of 20m/s at an angle of 30degree above the horizontal. Calculate 1) the time of flight to the ground 2) the velocity and impact

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Tài Đức 5 years 2021-08-31T09:32:12+00:00 1 Answers 25 views 0

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  1. Farah
    0
    2021-08-31T09:33:24+00:00
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    Answer:

    Explanation:

    u = initial velocity = 20m/s

    h = Height = 30m

    Theta = 30°

    g = acceleration due to gravity =9.8m/s^2

    Time of flight (T) = total time taken by an object reach its maximum height and return to the ground.

    T =( 2*u*sin(theta)) ÷g

    T = (2*20*sin(30°))÷9.8

    T = (40 * 0.5) ÷ 9.8

    T = 20 ÷9.8

    T = 2.04s

    Velocity (v)

    v = sqrt(2gh)

    v = sqrt(2*9.8*30)

    v = sqrt(588)

    v = 24.25m/s

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