Physics Help Forum take-off velocity

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Apr 2nd 2017, 06:39 PM #1 Junior Member   Join Date: Apr 2017 Posts: 2 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?
Apr 3rd 2017, 01:39 AM   #2
Senior Member

Join Date: Apr 2015
Location: Somerset, England
Posts: 479
 What do I do next?
Explain the problem in greater detail!

 Apr 3rd 2017, 05:09 AM #3 Senior Member   Join Date: Aug 2010 Posts: 151 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. topsquark likes this.
 Apr 3rd 2017, 06:48 AM #4 Junior Member   Join Date: Apr 2017 Posts: 2 I am supposed to calculate the vertical velocity at take-off. My two subjects had jumps of .254 m and .381 m.
 Apr 3rd 2017, 07:10 AM #5 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 479 Well HallsofIvy has given you a formula for each of those cases.

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