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Calculate the height of the center of mass above its starting height during a squat jump...

Calculate the height of the center of mass above its starting height during a squat jump based on the following information: BW = 777 N, total vertical force = 899 N, and time of force application = 0.93 seconds. The answer is 0.10 m, just not sure how to get there. Thank you in advance.

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Answer #1

net upward force = 899 - 777 = 122 N

impulse = 122 x 0.93 = 113.46 Ns

mv = 113.46 Ns => v = 113.46/(777/9.8) = 1.431

now h = v2/2g = (1.431)2/19.6 = 0.10448 m

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