Question
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
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.