Asked by Terry
If micheal Jordan has a vertical leap of 1.29 m, what is his take off speed and his hang time(total time to move upwards to the peak and then return to the ground)?
Answers
Answered by
Henry
a. V^2 = Vo^2 + 2g*h = 0
Vo^2 = -2g*h = -(-19.6)*1.29 = 25.284
Vo = 5.03 m/s
b. 0.5g*t^2 = 1.29 m.
4.9t^2 = 1.29
t^2 = 0.263
Tf = 0.513 s. = Fall Time.
Tr = Tf = 0.513 s. = Rise time.
Hang time = Tr + Tf = 0.513 + 0.513 = 1.026 s.
Vo^2 = -2g*h = -(-19.6)*1.29 = 25.284
Vo = 5.03 m/s
b. 0.5g*t^2 = 1.29 m.
4.9t^2 = 1.29
t^2 = 0.263
Tf = 0.513 s. = Fall Time.
Tr = Tf = 0.513 s. = Rise time.
Hang time = Tr + Tf = 0.513 + 0.513 = 1.026 s.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.