Calculating Vertical Jump Heights

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SUMMARY

The discussion focuses on calculating vertical jump heights, specifically addressing a 2-foot (0.6 m) vertical jump and the hang time associated with it. The time spent moving upward during the jump is calculated using the formula t=sqrt(2d/g), yielding 0.35 seconds, leading to a total hang time of 0.70 seconds. Additionally, the vertical height of Michael Jordan's jump is calculated based on a 1-second hang time, resulting in a height of 4.9 meters using the formula d=(1/2)gt². A correction was noted regarding the hang time calculation for part b.

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needhelp83
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Surprisingly, very few humans can jump more than 2 feet (0.6 m) straight up. Solve for the time one spends moving upward in a 2-foot vertical jump. Then double it for the "hang time" - the time one's feet are off the ground.
b) Calculate the vertical height of Michael Jordan's jump when he attains a hang time of a full 1 s.

a)
t=sqrt(2d/g) =sqrt(2(0.6 m))/9.8 =0.35 s

Hangtime=0.35 s x 2= 0.70 s

b)
d=(1/2)gt2=(1/2)(9.8 m/s2)(1.0 s)2= 4.9 m


Is this correct?
 
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needhelp83 said:
Surprisingly, very few humans can jump more than 2 feet (0.6 m) straight up. Solve for the time one spends moving upward in a 2-foot vertical jump. Then double it for the "hang time" - the time one's feet are off the ground.
b) Calculate the vertical height of Michael Jordan's jump when he attains a hang time of a full 1 s.

a)
t=sqrt(2d/g) =sqrt(2(0.6 m))/9.8 =0.35 s

Hangtime=0.35 s x 2= 0.70 s

b)
d=(1/2)gt2=(1/2)(9.8 m/s2)(1.0 s)2= 4.9 m


Is this correct?
you forgot to divide the hang time by 2 in part b. part a is a ok.
 
Last edited:

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