How high does the basketball player leap?

[Medgirl314]

Homework Statement

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?

Homework Equations

I think I could use y=y0t+v0t+1/2at^2

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.

Thanks in advance for clarification!

 

[CWatters]

You do know the final velocity.

Have a look at the other SUVAT notation here...
http://en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations

You know:

The final velocity v
The acceleration a
The displacement s

One of the SUVAT equations will give you the initial velocity u.

 

[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?

 

[CWatters]

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

 

[Medgirl314]

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

 

[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

 

[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!

 

[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?

 

[haruspex]

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?

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?

 

[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?

 

[haruspex]

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?

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 v_f. 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.

 

[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?

 

[haruspex]

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?

That'll work.

 

[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.

 

[haruspex]

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.

Why is that a problem?

 

[Medgirl314]

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

 

[haruspex]

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

Yes.

 

[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?

 

[haruspex]

Thanks!

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

Watch the signs!

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?

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.

 

[Medgirl314]

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

 

[haruspex]

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

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.

 

[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?

 

[Medgirl314]

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

 

[haruspex]

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

You now know, or have expressions for, a, s, v₀ and v_h (writing v_h 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.

 

[CWatters]

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?

See SUVAT or "Equations of Motion" on Wikipedia

 

[Medgirl314]

Thanks, both of you! Maybe posting five problems at a time is not the best idea. s=

s = v0t + at2/2
a=9.8 m/s^2
y=1.0
yinital=0 m

s=vot+9.8m/s^2/2

I'm sorry, I'm still a bit confused on the inital velocity. I thought we were still figuring that out. Thanks again!

 

[CWatters]

Try one of the other SUVAT equations..

s = Vt - 0.5 at²

You know

S = height
V = final velocity V=0
a = -g

It will simplify so no quadratic to solve.

 

[CWatters]

Duplicate post deleted

 

[Medgirl314]

Thanks, I didn't realize there were two.

s = Vt - 0.5 at
1=0-0.5(-9.8)t
1=4.9t

t=0.204 s

 

[CWatters]

Check your work..

t² not t

 

[Medgirl314]

s = Vt - 0.5 at2
1=0-0.5(-9.8)t^2
1=-0.5(-9.8)t^2
t=0.45 s

Thanks! Is that better?

 

[CWatters]

Looks good.

 

[Medgirl314]

Great, thanks again! So is this my final answer, or is that just the time it took for him to go *up*?

 

[Medgirl314]

Great, thanks again! So is this my final answer, or is that just the time it took for him to go *up*?

 

[CWatters]

Well spotted, I'd missed that. Yes just the time to go up. The time to come down will be the same.

 

[Medgirl314]

Great, thanks!