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A ball is thrown vertically upwards in the air at a velocity of 50(m/s) and reaches a maximum height of 20(m). Calculate the time when the ball reaches the 10(m) mark, while going up and coming down. Use the following equation of motion:

y(f) = y(i) + v(i)t+1/2at^2

y(f) and y(i) = final and initial position, v(i) = initial velocity, a = acceleration, and t = time, respectively.

  • Biomechanics -

    velocity is 50-9.8t.
    v=0 at top of trajectory, at t=50/9.8=5.10 sec.

    y = yi + 50t - 4.9t^2, so at t=5.10,

    20 = yi + 50(5.10) - 4.9(5.10)^2
    yi = -107.55
    eh? thrown from below ground?

    Yet it makes sense, since it was thrown upward with a velocity of 50 m.s, yet only made it 20m high. No mention was made of a horrific air resistance.

    Anyway, assuming the above,

    y = -107.55 + 50t - 4.9t^2
    solve for t when y=10:
    t = 3.67 or 6.53

    If I botched it, make the fix.

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