Posted by taylor on Thursday, January 5, 2012 at 4:39pm.
an olympic long jumper i capable of jumping 8.0 m. Assuming hi horizonal speed is 9.1 m/s as he leaves the ground, how long is he in the air and how high doe he go? Assume that he lands standing upright that is, the same way he left the ground.

physics  Henry, Friday, January 6, 2012 at 5:25pm
d = V*t,
t = d/V = 8 m / 9.1 m/s = 0.879 s.
Tr + Tf = 0.879 s,
Tr = Tf,
Tr + Tr = 0.879,
2Tr = 0.879,
Tr = 0.4395 s. = Rise time or time to reach max height.
Vf = Vo + gt.
Vo = Vf  gt,
Vo = 0  (9.8)(0.4395) = 4.307 m/s.
h = Vo*t + 0.5g*t^2,
h = 4.307*0.4395  4.9*(0.4395)^2 = 0.946 m.
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