Calculating Basketball Player's Hang Time

[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

I've been trying it with x=VoT + 1/2aT^2 but just can't get it right. Any help would be appreciated.

 

[daveed]

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

 

[needhelpperson]

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

I've been trying it with x=VoT + 1/2aT^2 but just can't get it right. Any help would be appreciated.

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.

 

[Webb]

so is .5 the correct answer?

 

[Locrian]

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

 

[Nenad]

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

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

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

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

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

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

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

$$t = 0.5s$$

Do you get it?