An Olympic long jumper is capable of jumping 8.3 m. Assume that his horizontal speed is 9.5 m/s as he leaves the ground. He was in the air for .87 seconds. How high does he go assuming that he lands upright, the same way he was when he left the ground?

Answer 1

8.3=viSinTheta*t you know vi, you know t, solve for theta.

you know it took .87/2 seconds to fall from max height.

h=1/2 g t^2 find h