How do I caculate hang time?

by mandaa123
Tags: caculate, formula, hang, homework, time
 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)
 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)
 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
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.
 P: 550 $$0.58 = \frac{1}{2} \times 9.81 \times t^2$$ Divide both sides by 1/2: $$\frac{0.58}{\frac{1}{2}} = 9.81 \times t^2$$ Divide both sides by 9.81: $$\frac{0.58}{\frac{1}{2} \times 9.81} = t^2$$ Since dividing by 1/2 is the same as multiplying by 2: $$\frac{2 \times 0.58}{9.81} = t^2$$ Taking the square root: $$\sqrt{\frac{2 \times 0.58}{9.81}} = t$$

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