An Olympic long jumper leaves the ground at an angle of 18.7 ° and travels through the air for a horizontal distance of 7.04 m before landing. What is the takeoff speed of the jumper?

Answer 1

Dx = Vo^2*sin(2A)/g = 7.04 m
Vo^2*sin(37.4)/9.8 = 7.04
Solve for Vo.