How do I caculate hang time?


by mandaa123
Tags: caculate, formula, hang, homework, time
mandaa123
mandaa123 is offline
#1
Sep26-08, 01:47 PM
P: 8
1. The problem statement, all variables and given/known data
Calculate the hang time of an athlete who jumps a vertical distance of 0.58 meter.



2. Relevant equations
all i know is that d= 0.5m, and possibly initial velocity is 0? im not sure.



3. The attempt at a solution
i tried using v = d/t, even though i doubted it would work.

(this homework is due today, i really need help)
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AshesToFeonix
AshesToFeonix is offline
#2
Sep26-08, 02:20 PM
P: 12
Total time (hang time) = time going up + time coming down.

and, time up = time down

so, 2 X time down = hang time.

for time down use the formula (yes V initial = 0, A = -9.81)

X final = X initial + V initial (t) + 1/2A(t^2)
mandaa123
mandaa123 is offline
#3
Sep26-08, 02:46 PM
P: 8
i understand what formula to use now, but im having trouble with the math because 0.5m=(-9.81Xt^2) / 2 and i do not know how to solve for t in that, since its squared, but over a fraction and multiplying with 9.81

AshesToFeonix
AshesToFeonix is offline
#4
Sep26-08, 04:40 PM
P: 12

How do I caculate hang time?


...well thats order of operations. you will have serious trouble passing without knowing them....

.58m = 0 + 0 + .5 (9.81m/s^2)(t^2)

to get t by itself

1) add or subtract from each side (in this case that part is 0)
2) multiply or divide
3) take your square root

t^2 = the sq root of (.58m / ((.5 times 9.81))

thats the time it takes to go down. doubling it will give you your total hang time.
Nick89
Nick89 is offline
#5
Sep26-08, 04:46 PM
P: 550
[tex]0.58 = \frac{1}{2} \times 9.81 \times t^2[/tex]
Divide both sides by 1/2:
[tex]\frac{0.58}{\frac{1}{2}} = 9.81 \times t^2[/tex]
Divide both sides by 9.81:
[tex]\frac{0.58}{\frac{1}{2} \times 9.81} = t^2[/tex]
Since dividing by 1/2 is the same as multiplying by 2:
[tex]\frac{2 \times 0.58}{9.81} = t^2[/tex]
Taking the square root:
[tex]\sqrt{\frac{2 \times 0.58}{9.81}} = t[/tex]


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