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3 June, 00:29

A baseball is thrown straight upward on the moon with an initial speed of 35 m/s.

Compute:

(a) the maximum height reached by the ball. > (b) the time taken to reach that height.

--> (c) its velocity 30 s after it is thrown.

(d) when the ball's height is 100 m.

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Answers (1)
  1. 3 June, 00:53
    0
    a) 376 m

    b) 21.5 s

    c) - 13.9 m/s

    d) 3.08 s, 39.9 s

    Explanation:

    Given:

    y₀ = 0 m

    v₀ = 35 m/s

    a = - 1.63 m/s²

    a) Find: y when v = 0 m/s.

    v² = v₀² + 2a (y - y₀)

    (0 m/s) ² = (35 m/s) ² + 2 (-1.63 m/s²) (y - 0 m)

    y = 376 m

    b) Find: t when v = 0 m/s.

    v = at + v₀

    0 m/s = (-1.63 m/s²) t + 35 m/s

    t = 21.5 s

    c) Find: v when t = 30 s.

    v = at + v₀

    v = (-1.63 m/s²) (30 s) + 35 m/s

    v = - 13.9 m/s

    d) Find: t when y = 100 m.

    y = y₀ + v₀ t + ½ at²

    100 m = 0 m + (35 m/s) t + ½ (-1.63 m/s²) t²

    100 = 35t - 0.815 t²

    Solve with quadratic formula:

    t = 3.08 s, 39.9 s
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