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Old Apr 2nd 2017, 07:39 PM   #1
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take-off velocity

I am trying to determine the take-off velocity of two vertical jumps. The displacement is 0.254 and 0.381. So far I have gotten this far.
sqrt(2 x 9.81 x 0.254)= 2.23 m/s
sqrt(2 x 9.81 x 0.381)= 2.73 m/s

What do I do next?
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Old Apr 3rd 2017, 02:39 AM   #2
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What do I do next?
Explain the problem in greater detail!

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Old Apr 3rd 2017, 06:09 AM   #3
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First, you should explain why you did that calculation! What formula were you using and how does it help solve this problem?

What I would do:
The distance traveled with constant negative acceleration g and initial speed v is d= -(g/2)t^2+ vt. We can "complete the square" by writing this as d= -(g/2)(t^2- (2v/g)t)= -(g/2)(t^2- (2v/g)t+ v^2/g^2- v^2/g^2)= -(g/2)(t- v/g)^2+ v^2/2g. Since a square is never negative, -(g/2)(t- v/g)^2 is never positive. The maximum height occurs when t- v/g= 0 and is v^2/2g.

Solve v^2/2g= 0.254 and v^2/2g= 0.381.
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Old Apr 3rd 2017, 07:48 AM   #4
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I am supposed to calculate the vertical velocity at take-off. My two subjects had jumps of .254 m and .381 m.
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Old Apr 3rd 2017, 08:10 AM   #5
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Well HallsofIvy has given you a formula for each of those cases.
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