A ball is thrown directly downward with an initial speed of $8.00 m / s$ from a height of $30.0 m$ . After what time interval does it strike the ground?

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Solution

The equation for the height of the ball as a function of time is
$y_{f} = y_{i} + v_{i} t - \frac{1}{2} g t^{2}$
$0 = 30 m + (- 8.00 m / s) t - (4.90 m / s^{2}) t^{2}$
Solving for $t$ ,
$t = \frac{+ 8.00 \pm \sqrt{(- 8.00)^{2} - 4 (- 4.90) (30)}}{2 (- 4.90)} = \frac{+ 8.00 \pm \sqrt{64 + 588}}{- 9.80}$
$t = 1.79 s$

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A ball is thrown directly downward with an initial speed of 8.00m/s from a height of 30.0m. After what time interval does it strike the ground?

The equation for the height of the ball as a function of time isyf=yi+vit−12gt20=30m+(−8.00m/s)t−(4.90m/s2)t2Solving for t,t=+8.00±√(−8.00)2−4(−4.90)(30)2(−4.90)=+8.00±√64+588−9.80t=1.79s

A ball is thrown directly downward with an initial speed of $8.00 m / s$ from a height of $30.0 m$ . After what time interval does it strike the ground?