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8 November, 14:22

David drops a ball from a bridge at an initial height of 70 meters. (a) What is the height of the ball to the nearest tenth of a meter exactly 2 seconds after he releases the ball? (b) How many seconds after the ball is released will it hit the ground?

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  1. 8 November, 15:16
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    A) y (2) = 70 - 4.9 * 2^2 = 70 - 19.6 = 50.4 m

    b) 0 = 70 - 4.9t^2

    70/4.9=t^2

    700/49 = t^2

    10 √ 7/7 = t (in seconds)

    t is approximately equal to 3.78 seconds
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