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A ball is thrown directly downward with an initial speed of 9.00 m/s, from a height of 29.8 m. After what time interval does it strike the ground?

Added Sun, 06 Sep '15

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y = 0m

y0 = 29.8m

v0 = -9 m/s

g = 9.8 m/s^2

t= ?

y=y0+v0t-0.5gt^2

0m = 29.8m + -9m(t) - 0.5(9.8)(t)^2

Quadratic Equation:

(9 +/- sqrt(9^2 - 4(-4.9)(29.8)))/2(-4.9)

t=1.7131s

Added Sun, 06 Sep '15

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