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

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

Thanks prior, all...

Homework Statement



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

Homework 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)

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!
 

Answers and Replies

  • #2
656
2


Thanks prior, all...

Homework Statement



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

Homework 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)

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!
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:
  • #3
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So at max height what is Vy and what is Vx in terms of Vo?

I am leading you to get everything in terms of Vo and then doing a little right triangle geometry. Remember at max height you are halfway through your trip and Vy = 0 m/s. If you use these facts, you can get the sides Vx (easy you already got it), and Vy (might be messy) in terms of Vo and then just take the inverse tangent... This is how I would go about it.
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:
  • #4
656
2


vmx = v0x
vmy = v0y/5

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?
See my new instructions. IM A bit tired and I was not thinking. Its straightforward.
 
  • #5
61
0


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...
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
656
2


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...
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.
velocity triangle.JPG


Unless I misunderstood the problem this is very straight forward. If you need further help post.
 
  • #7
61
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Haha, crap.

That's totally what I needed. Thanks for all your help!
 
  • #8
656
2


Haha, crap.

That's totally what I needed. Thanks for all your help!
Its sort of embarassing now that one looks back on it.
 
  • #9
61
0


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!
 

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