# Homework Help: How high does the basketball player leap?

1. Nov 8, 2013

### Medgirl314

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

From a stationary standing position, a basketball player leaps straight up, reaching a height og 1.0 m above the ground. How long is the player in the air?

2. Relevant equations

I think I could use y=y0t+v0t+1/2at^2
3. The attempt at a solution

I've found the formula that looks best to use for this problem. I know a=9.8 m/s^2. I know, letting up be positive, that y=1.0 m and yinital=0 m. The equation I want to use wants me to know the inital velocity. How do I find it? I don't think it is zero, because he's leaping up. I could use v^2=v0^2=2aΔx, but I don't know the final velocity, either.

2. Nov 8, 2013

3. Nov 8, 2013

### Medgirl314

Thank you!!

Okay, I get that the acceleration is 9.8 m/s^2, and the displacement is 1.0 m. Is the final velocity 0 since it he stops at his peak of the jump before falling?

4. Nov 10, 2013

### CWatters

Sorry for delay in replying. Yes. Final velocity at the top is zero (briefly!)

5. Nov 11, 2013

### Medgirl314

Thank you! The delay is quite alright.So I can use 0 as the final velocity for my entire equation?

6. Nov 11, 2013

### CWatters

I'm more familiar with it written..

s = vt - 0.5at2

where
s = displacement (height) = 1m
v = final velocity = 0
a = acceleration = -g
t = time

7. Nov 11, 2013

### Medgirl314

Okay, thanks! I think there's a compromise here. I only know it well the other way, so I'll work it out, and you can tell me if that step was right. Thanks again!

8. Nov 11, 2013

### Medgirl314

If it doesn't add complications, I think I'll let up be positive.
y=y0t+v0t+1/2at^2
y=1/2 at^2
1=1/2 (9.8*t)

1=4.9*t

t=0.204 s.

Is that right? Would I need to double that answer?

9. Nov 11, 2013

### haruspex

v0 usually denotes initial velocity, and indeed that is the form of equation you've written. But you don't know v0. However, you can reverse the problem and ask how long it takes to fall from that height, starting at rest. This is effectively what you have done.
Even so, there is an error. What happened to the power of 2 on t?

10. Nov 11, 2013

### Medgirl314

Oh, I forgot to square it. Thank you! I thought the inital velocity was 0, since before he jumped, he wasn't moving. Is it the rate at which he jumped, instead?
y=y0t+v0t+1/2at^2
1=v0t+1/2(9.8)^2

Now do I need a different equation to find the inital velocity?

11. Nov 11, 2013

### haruspex

Yes, it's the launch speed. If you set v0=0 and apply the formula the distance you get will be how far you'd fall in that time. As I wrote, that will answer the question, simply by thinking about the process in reverse. But if you want to do it by standard methods then you need an equation, or equations, involving s, t, a and vf. CWatters quoted such. If you want to get there using the equations you know, you will need to use two of them and eliminate v0 between them.

12. Nov 11, 2013

### Medgirl314

Okay, thank you! So you're basically saying I need another equation to find the inital velocity, then I can use my equation to find the time, correct?

13. Nov 11, 2013

### haruspex

That'll work.

14. Nov 11, 2013

### Medgirl314

Okay, thanks! Would v^2=v0^2+2aΔx would work? I think I know all the needed information to solve, but since v^2 equals 0, the equation doesn't seem to be in the right format.

15. Nov 12, 2013

### haruspex

Why is that a problem?

16. Nov 12, 2013

### Medgirl314

Oh, I just now realized it is a quadriatic equation. Correct?

17. Nov 12, 2013

### haruspex

Yes.

18. Nov 12, 2013

### Medgirl314

Thanks!

0=v0^2+2(9.8)(1)

0=v0^2+19.6

Now that I simplify, it looks like I only have three terms instead of four. Am I suppossed to forget about simplifying until the end?

19. Nov 12, 2013

### haruspex

Watch the signs!
As I wrote in post #11, you want two equations involving v0 and eliminate v0 between them. It's always best to leave plugging in numbers to the final step. It makes it easier to spot mistakes, easier for others to follow, and minimises accumulation of rounding errors.

20. Nov 12, 2013

### Medgirl314

Okay, so this is a quadriatic equation, but I *don't* want to use the quadriatic formula to solve, correct? Thanks again!

21. Nov 12, 2013

### haruspex

Well, you will have to take a square root at some point for this question, no matter which way you proceed. But I would not calculate v0. Continue with the algebra until you have an expression for time in terms of all the given quantities.

22. Nov 12, 2013

### Medgirl314

Thanks!
Then I need a new equation, right? Since the following one doesn't involve time? v^2=v0^2+2aΔx Am I keepoing that equation for future use, but now using an equation for time?

23. Nov 18, 2013

### Medgirl314

I can't edit that post, unfornutaley. I meant keeping. What equation should I used for this step? Thanks!

24. Nov 18, 2013

### haruspex

You now know, or have expressions for, a, s, v0 and vh (writing vh for the velocity at highest point, i.e. 0). You can use any equation involving three of those and t to get the time to highest point. Any SUVAT equation except the one you already used will do that.
Or you can switch to considering the whole trajectory. You have initial speed, acceleration and distance (which will be what on landing?), so you can use s = v0t + at2/2 to get the total time in the air.

25. Nov 19, 2013

### CWatters

See SUVAT or "Equations of Motion" on Wikipedia