1. Aug 31, 2004

### Webb

A basketball player jumps 1 meter high off the ground, turns around and starts back down.

Estimate the time she is within 30 cm of the top of her trajectory (her hang time.)

(HINT: Calculate the time it takes to fall 30 cm from rest and double it.)

Explain why that works

note g=9.8

Ive been trying it with x=VoT + 1/2aT^2 but just cant get it right. Any help would be appreciated.

2. Aug 31, 2004

### daveed

i dont get why you wouldn't get the right answer. whats the correct answer?
did you remember to write 30 cm as .3m?

3. Aug 31, 2004

### needhelpperson

it works because the trajectory is symetrical, thus you only need to find out the time for half the trajectory, and multiply it by 2. As well, you're treating the trajectory as if the person only jumped 30 cm high, which is a lot simpler than an alternate method, since you're just finding the necessary time without having to find any excess information.

Last edited: Aug 31, 2004
4. Aug 31, 2004

### Webb

so is .5 the correct answer?

5. Aug 31, 2004

### Locrian

Only if you put the correct units on it :tongue2:

6. Aug 31, 2004

$$d = v_{o} t + \frac{1}{2}at^2$$

$$-0.3m = \frac{1}{2}(-9.8m/s^{2})t^2$$

$$-0.6m = (-9.8m/s^{2})t^2$$

$$\frac{-0.6m}{-9.8m/s^{2}} = t^2$$

$$0.0612s^2 = t^2$$

$$\sqrt{0.0612s^2} = \sqrt{t^2}$$

$$t = (0.25s)(2)$$

$$t = 0.5s$$

Do you get it?