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

[Beamsbox]

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,
v_y = v₀ sin theta₀ - gt,
v_y² = (v₀ sin theta₀)² - 2g(y - y₀)

The Attempt at a Solution

So, I've drawn the path, defining the maximum heigth as v_m.
At v_m, the vertical component, v_my, must be 0.
v_mx must be constant and would equal the same as it started,
v_mx = v_ox

In the problem, the initial velocity, v₀ is 5 times v_m, so
v₀ = 5(v_m) = 5(v_ox)

Not sure where to go from here, I tried to break v_o into its components, but just ended up with v_oy² = 24v_ox²


Any ideas?

Thanks again!

 

[pgardn]

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,
v_y = v₀ sin theta₀ - gt,
v_y² = (v₀ sin theta₀)² - 2g(y - y₀)

The Attempt at a Solution

So, I've drawn the path, defining the maximum heigth as v_m.
At v_m, the vertical component, v_my, must be 0.
v_mx must be constant and would equal the same as it started,
v_mx = v_ox

In the problem, the initial velocity, v₀ is 5 times v_m, so
v₀ = 5(v_m) = 5(v_ox)

Not sure where to go from here, I tried to break v_o into its components, but just ended up with v_oy² = 24v_ox²


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... I am tired, but yes, I think its this easy. Vx does not change...

 

[Beamsbox]

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.

v_mx = v_0x
v_my = v_0y/5 = 0

This is just an assumption, based on the thought that if v_x remains the same, then the only way for the initial velocity to be 5x v_m, then it's v_y that must be 5x, so the inverse of that would be that at v_m, v_my must be the the one that changes, therefore, 1/5^th of the starting (v_0y) velocity...

Is this correct?

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

 

[pgardn]

v_mx = v_0x
v_my = v_0y/5

This is just an assumption, based on the thought that if v_x remains the same, then the only way for the initial velocity to be 5x v_m, then it's v_y that must be 5x, so the inverse of that would be that at v_m, v_my must be the the one that changes, therefore, 1/5^th of the starting (v_0y) velocity...

Is this correct?

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

 

[Beamsbox]

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... I am 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. (v_ox² + v_oy²) = 5(v_m)

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

v₀ = sq.rt. (v_ox² + v_oy²) = 5(v_ox)
v_ox² + v_oy² = 5²(v_ox)²
v_oy² = 24v_ox²
(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...

 

[pgardn]

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. (v_ox² + v_oy²) = 5(v_m)

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

v₀ = sq.rt. (v_ox² + v_oy²) = 5(v_ox)
v_ox² + v_oy² = 5²(v_ox)²
v_oy² = 24v_ox²
(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 don't 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.

 

[Beamsbox]

Haha, crap.

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

 

[pgardn]

Haha, crap.

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

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

 

[Beamsbox]

ha ya, I know...

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