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Ryan throws a tennis ball straight up into the air. The ball reaches its maximum height at 2 seconds. The approximate height of the ball x seconds after being thrown is shown in the table.

Which equation models the motion of the ball?

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2 answers
  1. well, you know that

    y = vt - 4.9t^2

    max height is at t = v/9.8

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  2. let the initial velocity be k m/s
    so
    height = -4.9t^2 + kt
    d(height)/dt = -9.8t + k
    when t = 2, d(height)/dt = 0
    2 = -9.8(2) + k
    k = 19.6

    so your equation was
    height = -4.9t^2 + 19.6t
    the maximum height is
    = -4.9(4) + 19.6(2) m
    = appr 19.6 m

    I used the more traditional variable t for time instead of the suggested x, no big deal.

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