The formula for the vertical height, h, attained with a vertical take off speed of u is u² - 2gh = 0 where g = 9.8 m/s², acceleration due to gravity.
Because h = 1.29 m, therefore u² = 2*9.8*1.39 = 27.244 u = 5.2196 m/s
Also, the time of flight, t, required when the vertical take-off speed is u is given by ut - 0.5gt² = 0. Therefore 5.2196t - 0.5*9.8*t² = 0 5.2196t - 4.9t² = 0 t(5.2196 - 4.9t) = 0 t = 0, or t = 5.2196/4.9 = 1.0652 s
t = 0 corresponds to take-off. t = 1.0652 s corresponds to landing. The hang time is 1.0652 s.
Answer: Take-off speed = 5.22 m/s (nearest hundredth) Hang time = 1.065 s (nearest thousandth)
