
#1
Sep2608, 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) 



#2
Sep2608, 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) 



#3
Sep2608, 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




#4
Sep2608, 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. 



#5
Sep2608, 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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