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April 18, 2014

April 18, 2014

Posted by **Kristen** on Sunday, September 9, 2007 at 10:55pm.

- physics -
**drwls**, Sunday, September 9, 2007 at 11:47pmIf he jumps up H = 0.62 m, the initial velocity V is sucn that

(1/2) M V^2 = M g H

V = sqrt (2gH) = 3.48 m/s

Height vs time is given by

y = 3.38 t - 4.90 t^2

Solve for the times when H = 0.15 m and y = 0.47 m. You will need to use the postive root of the quadratic eauation.

At the top of the trajectory, H = 0.62 m and g t = V = 3.48 m/s, so t = 0.355 s there.

t = 0 when y = 0. Knowing the times when H = 0.15 and 0.47 m will let you calculate the intervals they are asking for.

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