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December 17, 2014

December 17, 2014

Posted by **Anon14** on Sunday, September 11, 2011 at 10:58pm.

for the first rock i used:

t(ground)=square root((2yo)/g)

... then I subtracted 2.7s from that number

I think you use Vo=(y-0)+.5gt^2/t

...I am not getting the correct answer. Any advice?

- College Physics -
**drwls**, Sunday, September 11, 2011 at 11:23pmI agree that the dropped rock takes sqrt(2Yo/g) = 6.03 s to teach the ground. The thrown rock must take 6.03 - 2.7 = 3.33 s from the timne it is thrown.

178.2 = Vo*t - (g/2)t^2

Solve for Vo.

Vo = (178.2 + 4.9 t^2)/t

With parentheses in the right place, and t = 3.33 s, you should get the right answer.

- College Physics -
**Anon14**, Sunday, September 11, 2011 at 11:31pmdoes my answer need to be negative due to the rocks falling?

- High School -
**Precalla**, Tuesday, October 4, 2011 at 11:59am-37.19 m/s

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