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22 September, 12:17

A ball is thrown directly downward with an initial speed of 8.50 m/s, from a height of 29.7 m. After what time interval does it strike the ground?

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  1. 22 September, 16:03
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    Using the following formula for linear-motion, the missing variable can be solved:

    s = Vi * t + 1/2 (a * t^2)

    Where: s = displacement = 29.7 m

    Vi = initial velocity = 8.5 m/s

    a = acceleration = 9.8

    t = time = ?

    Substituting:

    29.7 = 8.5t + 1/2 (9.8*t^2)

    29.7 = 8.5t + 4.9t^2

    Dividing both sides by 4.9:

    6.06 = 1.73t + t^2

    t^2 + 1.73t - 6.06 = 0

    (t - 1.74) (t + 3.48) = 0

    t = 1.74s

    From the above values, the correct answer is 1.74 seconds.
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