A good athlete can high jump 3m. What is the minimum speed an athlete will need to reach this height.?

A good athlete can high jump 3m. What is the minimum speed an athlete will need to reach this height. Where gravity is 10 m/s|||I think you meant the acceleration due to gravity = 10 m/s², and the first guy is right. the second guy didn't round correctly or has round off error, but his obtuse work is correct.





I'm assuming this is a question from a high school physics class. This is a "classic" Conservation of Mechanical Energy problem, as long as you neglect non-conservativee forces like wind resistance. =%26gt; K₁ + U₁ = K₂ +U₂





Event 1 is when the athlete's is standing. Event 2 is when the athlete's center of mass rose 3ft.





let's call


the mass of the athlete M,


v₁ the speed at event 1, which is what we're trying to solve for,


v₂ the speed at event 2, which would be zero when v₁ is at a minimum,


h₁ the initial height,


h₂ the final height the athlete's center of mass will reach, which for this problem is 3m.








K₁ = ½M v₁² (The kinetic energy of the athlete before leaving the ground)





K₂ = ½M v₂² =0 (The kinetic energy of the athlete at h2)





U₁ = M g h₁ = 0 (The potential energy of athlete can be set to zero at one event. I chose Event 1)





U₂ = M×g× h₂ where g = 10m/s² and h = 3m





now let's plug these into our equation


****** K₁ + U₁ = K₂ +U₂


½ M v₁² + 0 = 0 + M g h





Divide M from both sides, solve for v1


*** v₁ = √ (2g h₂)


*** v₁ = √ (2 ×10m/s² × 3 m)


*** v₁ = 7.75 m/s|||vf^2 = vi^2 + 2ad


0 = vi^2 + 2(-10m/s^2)(3m)


60m^2/s^2 = vi^2


7.75m/s = vi|||I'm assuming that this is a vertical leap?





Direct answer --%26gt; 7.74 m/s My math is below, but I checked this against an online calculator, and it's all correct.





Given values:





H = 3 m


g = 10 m/s^2


θ° = 90





Since we know the height, we can calculate the time that will take you come down. Since it's a vertical leap, that will be the same as the time to come up. Just a different equation.





Tf = SQRT [ 2H / g ]





Tf = SQRT [ (2 * 3m) / (10 m/s^2) ]


Tf = SQRT [ (6 m) / (10 m/s^2) ]


Tf = SQRT [ 0.6 s^2 ]


Tf = 0.775 s





Now, given the height at the beginning, and knowing the time that it takes to get up to the top, we can calculate the initial push-off velocity that the jumper has to have.





H = VoTf - 0.5gTf^2





(3 m) = (Vo) * (0.775 s) - [ 0.5 * (10 m/s^2) * (0.775 s)^2 ]


(3 m) = (Vo) * (0.775 s) - [ (5 m/s^2) * (0.60 s^2) ]


(3 m) = (Vo) * (0.775 s) - [ 3 m ]


(6 m) = (Vo) * (0.775 s)





Vo = 7.74 m/s|||2gh = v^2





2 x 10 x 3 = v^2 = 60, sq-rt = v = 7.746 m/s