How to Calculate Maximum Jump Height with Given Variables

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To calculate the maximum jump height of a 73 kg high jumper with an initial vertical velocity of 7.6 m/s, the relevant energy conservation equation is mgh1 + 1/2mv1^2 = mgh2 + 1/2mv2^2. At the peak of the jump, the final velocity (v2) is zero, simplifying the equation to 0 + 1/2(7.6)^2 = 9.8(h2). This allows for the calculation of height (h2) by solving for it after substituting the known values. The discussion clarifies that the initial velocity (v1) applies only at the start, while the final velocity (v2) is zero at the maximum height. The approach confirms the correct application of the energy conservation principle in solving for jump height.
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Homework Statement



A 73 kg high jumper leaves the ground with a vertical velocity of 7.6 m/s.
How high can he jump? The acceleration of gravity is 9.8 m/s^2.


Homework Equations



mgh1 + 1/2mv1^2 = mgh2 + 1/2mv2^2

The Attempt at a Solution



0 + 1/2(7.6)^2 = 9.8(h2) + 1/2(v2)^2

How can you solve this when you two unknowns?
Does v1 = 7.6 also apply to v2?
Am I using the right equation?
 
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Correct equation...Concept:when the jumper has jumped his highest at that point "instantaneously" his velocity becomes zero..So set v2=0 and head on with the math.
 
Alright thank you, that helps a ton.
 

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