Calculate the hang time of an athlete who jumps a vertical distance of 0.75 meter.?

Question

Please help me and show me how to do this problem!

Answer 1

The time going up will equal the time returning:
s = s0 + v0t + (1/2)at^2
Takling the top of the jump as 0 and positive direction downward,
h = 0 + 0 + (1/2)gt^2
t = √(2h/g)
the hang time will be twice the time to rise or fall:
T = 2√(2*0.75/9.80665)
T ≈ 0.78

Answer 2

Vertical Jump Meter

Answer 3

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Answer 4

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Answer 5

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RE:
Calculate the hang time of an athlete who jumps a vertical distance of 0.75 meter.?
Please help me and show me how to do this problem!

Answer 6

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Answer 7

You need to split this into two halves. The time to go up to the top, and the time to go down from there back to the ground. Each half will take the same amount of time, so long as we ignore complications like air resistance (which it's safe to say we can for this problem).

The second half, you have enough information for. You have an initial velocity of zero, a distance traveled of 0.75m, and an acceleration downward of 9.8 m/s^2. From this, you can find the time to go from the midpoint down, then just double it for the total "hang time".

Answer 8

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