1. Jan 2, 2012

### Casey314stl

1. The problem statement, all variables and given/known data

A basketball player achieves a hang time of 0.978 s in dunking the ball.
What vertical height will he attain? The
acceleration of gravity is 9.8 m/s^2

2. Relevant equations
?
3. The attempt at a solution 9.8/0.978^2= 10.245
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Jan 2, 2012
2. Jan 2, 2012

### DaveC426913

Yes. You can.

We do not (can not) do homework for you. Show your work, we can help.

3. Jan 2, 2012

### Casey314stl

The attempt at a solution 9.8/0.978^2= 10.245

4. Jan 2, 2012

### mtayab1994

Remember that H=.5gt^2

Retry and see what you get.

5. Jan 2, 2012

### Casey314stl

I tried H=.5*g*t^2 and got 4.68677?

6. Jan 2, 2012

22.9653?

7. Jan 2, 2012

### mtayab1994

Remember that time going up = time coming down .

8. Jan 2, 2012

### mtayab1994

No since t(up)=t(down)

then: H=.5g*(t/2)^2

9. Jan 2, 2012

### Casey314stl

what?

10. Jan 2, 2012

### Casey314stl

Thanks a lot I got it but why is it half time^2?

11. Jan 2, 2012

### mtayab1994

When you have the time going up = time going down, then you have: t(up)=t(down)=.978/2

I'm helping you too much.

12. Jan 2, 2012

### Casey314stl

I already got the the answer from the last response I was just wonder why it was half the time^2 if that was for every problem for this equation or just if an object is traveling up?

13. Jan 2, 2012

### mtayab1994

Well since he is jumping upwards he only takes half of the time to get to his maximum height. Therefore the rest of the time is spent coming down. Hence no need for the time coming down.