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11 April, 12:55

a ball is thrown vertically upward with an initial speed of 40 m/s. how high is the ball above the ground when it stops

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Answers (2)
  1. 11 April, 14:32
    0
    80m

    Explanation:

    According to the equation of motion, v² = u²+2as

    Where v is the final velocity of the body at maximum height = 0m/s

    u is the initial velocity of the body = 40m/s

    a = - g (negative acceleration due to gravity since the object is thrown upward) = - 10m/s²

    S = H = the height of the ball above the ground

    Substituting the values into the equation given, we will have;

    0² = 40²-2 (10) H

    0 = 1600-20H

    -1600 = - 20H

    H = - 1600/-20

    H = 80m

    The height of the ball above the ground where it stops will be 80m
  2. 11 April, 14:44
    0
    80m, assuming g=10m/s^2

    Explanation:

    40m/s will be reduced to 0m/s in 4 seconds. 4 seconds x 40m/s would be 160m up, but you will only get half of that because you decelerate linearly to 0m/s. This leaves you with 4 x 20 = 80m.
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