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pyhsics help i am totally lost?
Surprisingly, very few athletes can jump more than 2.6 ft (0.78 m) straight up. Use d = 1/2 gt2 and solve for the time one spends moving upward in a 2.6 foot vertical jump. Then double it for the "hang-time" -- the time one's feet are off the ground.
can someone explain this to me
2 Answers
- NoneLv 71 decade agoFavorite Answer
Certainly!
The given expression comes from one of the equations of motion:
http://en.wikipedia.org/wiki/Equations_of_motion
d = gt^2/2
g, the acceleration due to gravity at the Earth's' surface = 32 ft/sec^2
2.6 = 16t^2
t^2 = 0.1625
t = 0.403 sec
Exactly the same amount of time is required to come down again, so hang time is 0.806 sec. or 0.81 sec rounded
- Old Science GuyLv 71 decade ago
you are given everything needed to solve this
the distance equation
d = Vi*t + 1/2*g*t^2
where d=2.6, Vi=0 (standing start) and g= (-32.2)
simplifying
t = sqrt(2*d/g)
you double this time because it takes the same amount of time to come down as it took to go up - g is constant
