High Jump Physics: Leaping 1.55m with 0.70m/s Speed

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In the high jump, the transformation of kinetic energy into gravitational potential energy is crucial for achieving height. To determine the minimum speed required for an athlete to clear a height of 1.55 meters while leaving the ground with a speed of 0.70 m/s, the kinetic energy equation is applied. The initial kinetic energy must equal the sum of the potential energy at the peak and the kinetic energy during the jump. A calculation error was identified, specifically in the square root operation, leading to an incorrect speed result. Correcting this mistake is essential for accurately determining the necessary takeoff speed.
Morgan89
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In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must the athlete leave the ground in order to lift his center of mass 1.55 m and cross the bar with a speed of 0.70 m/s


I have tried solving this problem by using the formulas for kinetic and potential energy. I thought that the initial kinetic energy would equal the potential energy plus the kinetic energy in the air.

KE = PE + KE'
.5(m)(v^2) = m(g)(h) + .5(m)(v^2)
Then i figured i could cancel out mass since it is in all parts
.5(v^2) = (9.81)(1.55) + .5(.7^2)
v= Sqrt (15.205 + .245)
v = 30.9

This is the wrong answer, and i am not sure what i am doing wrong. Please help.
 
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Did you actually take the square root? It doesn't look like you did.
 
Thank you. that was the problem! Appreciate it!
 

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