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9 January, 22:21

Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 2 seconds.

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  1. 9 January, 22:34
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    Answer:.

    Required velocity = 6.26ms^-1

    Explanation:

    Given,

    Distance, s = 450m

    Time, t = 2 sec

    Step 1. We obtain the distance covered within the given time under gravitational acceleration, g = 9.8ms^-2

    S = ut + (1/2) gt^2. :; u = 0

    : S = (1/2) gt^2

    = (1/2) (9.8) (2^2)

    = 19.6m

    Step 2:

    We obtain the velocity using the formula.

    V^2 = u^2 + 2gs.

    Where u is initial velocity, v is final / required velocity

    Again u = 0

    : V^2 = 2 (9.8) (19.6)

    = 39.2

    : V = 6.26ms^-1
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