Calculate the hang time of an athlete who jumps a vertical distance of 0.76 meter.
what would i use as equation
what would i use as equation
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do it the opposite way, how long does an object take to fall from that height, then double it.
t = 2 x √(2h/g) = 2√(2•0.76/9.8)
t = 2 x √(2h/g) = 2√(2•0.76/9.8)
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d = 1/2at^2
(.76m) = 1/2 (9.81m/s^2)t^2
t= .39 sec
Since its vertical, the distance is .76m, when hes up their, the acceleration is 9.81 m / s^2 (gravity) and solve for t.
(.76m) = 1/2 (9.81m/s^2)t^2
t= .39 sec
Since its vertical, the distance is .76m, when hes up their, the acceleration is 9.81 m / s^2 (gravity) and solve for t.