Simplifying:t \approx 0.34 secondsCalculating Hang Time for Vertical Jump

[mandaa123]

Homework Statement

Calculate the hang time of an athlete who jumps a vertical distance of 0.58 meter.

Homework Equations

all i know is that d= 0.5m, and possibly initial velocity is 0? I am not sure.

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)

 

[AshesToFeonix]

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]

i understand what formula to use now, but I am 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]

...well that's 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]

$$0.58 = \frac{1}{2} \times 9.81 \times t^2$$
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:
$$\sqrt{\frac{2 \times 0.58}{9.81}} = t$$