# Projectile Motion - V0 is 5 times v @ max height?

1. Jun 6, 2010

### Beamsbox

Projectile Motion - V0 is 5 times v @ max height?!?

Thanks prior, all...

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

A projectile's launch speed is 5 times its speed at maximum heigth. Find launch angle theta0.

2. Relevant equations
Not sure which are relevant on this one, but the given for the section are:

x - x0 = (v0 cos theta0)t,
y - y0 = (v0 sin theta0)t - 1/2gt2,
vy = v0 sin theta0 - gt,
vy2 = (v0 sin theta0)2 - 2g(y - y0)

3. The attempt at a solution

So, I've drawn the path, defining the maximum heigth as vm.
At vm, the vertical component, vmy, must be 0.
vmx must be constant and would equal the same as it started,
vmx = vox

In the problem, the initial velocity, v0 is 5 times vm, so
v0 = 5(vm) = 5(vox)

Not sure where to go from here, I tried to break vo into its components, but just ended up with voy2 = 24vox2

Any ideas?

Thanks again!

2. Jun 6, 2010

### pgardn

Re: Projectile Motion - V0 is 5 times v @ max height?!?

Much easier than all this.

Look at your right triangle of velocities initially. You know what the hyp is... Vo. You have already stated what Vx is in terms of Vo... Use your trig. Cosine... Im tired, but yes, I think its this easy. Vx does not change...

Last edited: Jun 6, 2010
3. Jun 6, 2010

### Beamsbox

Re: Projectile Motion - V0 is 5 times v @ max height?!?

vmx = v0x
vmy = v0y/5 = 0

This is just an assumption, based on the thought that if vx remains the same, then the only way for the initial velocity to be 5x vm, then it's vy that must be 5x, so the inverse of that would be that at vm, vmy must be the the one that changes, therefore, 1/5th of the starting (v0y) velocity...

Is this correct?

(I think I'm just confusing myself with this one...)

Last edited: Jun 6, 2010
4. Jun 6, 2010

### pgardn

Re: Projectile Motion - V0 is 5 times v @ max height?!?

See my new instructions. IM A bit tired and I was not thinking. Its straightforward.

5. Jun 6, 2010

### Beamsbox

Re: Projectile Motion - V0 is 5 times v @ max height?!?

No worries, I'm a bit tired as well.

(I kind of did that earlier, but got lost somewhere, I suppose.)

Here's what I was thinking...

The initial triangle broken down:

v0 = sq.rt. (vox2 + voy2) = 5(vm)

And since the velocity at max height, is ONLY the x component, and hence the same as the initial x component, (vm = vmx = vox), then you should be able to substitute vox for vm, leaving you with:

v0 = sq.rt. (vox2 + voy2) = 5(vox)
vox2 + voy2 = 52(vox)2
voy2 = 24vox2
(Which is what I posted in my first post...)

I'm not sure where the angle comes in, which formula that has theta in it to use...

6. Jun 7, 2010

### pgardn

Re: Projectile Motion - V0 is 5 times v @ max height?!?

The velocity in the x-direction remains the same the entire trip. And we know what that is. You basically stated it was 1/5 of its original speed Vo. Look at my diagram. And realize you know what cosine theta is from the diagram. Solve for theta. We made it way too difficult.
You dont need to use the pythagoream theorem. Cosine theta = adjacent side/ hypot.

Unless I misunderstood the problem this is very straight forward. If you need further help post.

7. Jun 7, 2010

### Beamsbox

Re: Projectile Motion - V0 is 5 times v @ max height?!?

Haha, crap.

That's totally what I needed. Thanks for all your help!

8. Jun 7, 2010

### pgardn

Re: Projectile Motion - V0 is 5 times v @ max height?!?

Its sort of embarassing now that one looks back on it.

9. Jun 7, 2010

### Beamsbox

Re: Projectile Motion - V0 is 5 times v @ max height?!?

ha ya, I know...

I kept looking at the vy as voy, so I guess I just wasn't making the association... can't say I won't do that again... Cheers!