- #1
Morgan89
- 9
- 0
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.
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.