Projectile motion with unknown angle and Vo

[Femme_physics]

Homework Statement

http://img854.imageshack.us/img854/6055/footballer01.jpg

A footballer kicks a ball and therefor the ball arrives at point C, above the goal at 4 meters height

Presuming that point C is at max height of the path and distance AB is 20 meters, calculate the speed of the ball (Va - angle and magnitude) during the kick. Presume that the ball is a material point.

The Attempt at a Solution

I'm getting stuck. I'm not sure whether it's my issue with solving trigonometric equations, or have I not applied the correct method? This is the first time I find a projectile motion problem to be so tough!

http://img8.imageshack.us/img8/962/footballer02.jpg

 

[I like Serena]

You're mixing up your symbols.

In equation #2 you're using v₀, while this should be v_y.
That should make it (quite a bit) easier. ;)

EDIT: Btw, where did you get equation #4 from?

 

[Femme_physics]

You're mixing up your symbols.

In equation #2 you're using v0, while this should be vy.
That should make it (quite a bit) easier. ;)

Hmm. I'll try it now then.
Geez...what a tough problem!

EDIT: Btw, where did you get equation #4 from?

This?

http://img852.imageshack.us/img852/3820/thisr.jpg

Uploaded with ImageShack.us

Our lecturer gave it to us. He seemed to be holding it as a really special great formula!

 

[Quinzio]

The movement of the ball can be split in two indipendent components: one vertical and one horizontal.
Then you can mix them up in the end.

In addition you may know that almost every law is reversible in time: that means you can imagine someone throwing a ball at the top of the 4 meters pole and the ball has to reach the player.

 

[Femme_physics]

In addition you may know that almost every law is reversible in time: that means you can imagine someone throwing a ball at the top of the 4 meters pole and the ball has to reach the player.

I don't appear to have a need for time. Adding time into the equations seem like extra work!

From what you're saying though, it appears that I'm making my life harder using the trajectory formula. Can I simply use the formulas of


http://img802.imageshack.us/img802/8066/3components.jpg


Wait, ignore that last equation. It should be V²=Vx²+Vy²

 

[I like Serena]

This?

Our lecturer gave it to us. He seemed to be holding it as a really special great formula!

Ah, I see.
I think it should be:

$$g \cdot h = \frac {v_0^2 \cdot \sin^2 \alpha} 2$$

That should give you the answer as well.

 

[Femme_physics]

Just to keep tab, I got 3 unknowns (angle, vox and voy) and I can use 5 equations to try and solve it?

*also, I'm surprised he confused the formula! Or maybe I mis-copied it? Hmm...

 

[Quinzio]

I don't appear to have a need for time. Adding time into the equations seem like extra work!

From what you're saying though, it appears that I'm making my life harder using the trajectory formula. Can I simply use the formulas of


http://img802.imageshack.us/img802/8066/3components.jpg


Wait, ignore that last equation. It should be V²=Vx²+Vy²

Ok, and how much time take for an object to fall from an height H under the acceleration of gravity g ?

 

[Femme_physics]

I just realized I accidentally added "time" into the equation

 

[I like Serena]

Just to keep tab, I got 3 unknowns (angle, vox and voy) and I can use 5 equations to try and solve it?

*also, I'm surprised he confused the formula! Or maybe I mis-copied it? Hmm...

Hmm, those other formulas are not quite right either.

It should be:

$$v_x=v_0 \cos \alpha$$
$$v_y=v_0 \sin \alpha - g \cdot t$$or

$$v_{0x}=v_0 \cos \alpha$$
$$v_{0y}=v_0 \sin \alpha$$

 

[Femme_physics]

Ok, and how much time take for an object to fall from an height H under the acceleration of gravity g ?

http://img696.imageshack.us/img696/6452/tttgr.jpg

Uploaded with ImageShack.us

But is that the same time it takes to go from the ground up?

 

[I like Serena]

http://img696.imageshack.us/img696/6452/tttgr.jpg

Uploaded with ImageShack.us

But is that the same time it takes to go from the ground up?

Yes!

So how much distance did it cover horizontally in that time?

 

[I like Serena]

Just to keep tab, I got 3 unknowns (angle, vox and voy) and I can use 5 equations to try and solve it?

*also, I'm surprised he confused the formula! Or maybe I mis-copied it? Hmm...

Actually you have a 4th unknown: v0.

And yes, you can try and solve it.

I can also suggest another approach.
Split the initial velocity in a horizontal and a vertical component, and set up the equations for the horizontal and vertical distance and speed.

Up to you though.

 

[Femme_physics]

That 4th unknown is messing up my solution. I'm stuck!

(I also used the result I got for "t" in those equations)

http://img580.imageshack.us/img580/672/2eq2un.jpg

 

[Quinzio]

Before you were able to state that the ball takes 0.903 sec to reach the top.

At the top you also know the ball doesn't rise further, so Vy must be... ?
You know g, t, Vy when the ball is at the top, get a value for $$V_0 \ sin (\alpha)$$
using your equations.

 

[Femme_physics]

http://img542.imageshack.us/img542/6445/voxs.jpg

http://img695.imageshack.us/img695/4441/solvedyay.jpg

Got it :)

So easy after I got "time". There weren't even 3 eq. and 3 unknown. Not even 2 eq 2 unknowns! Just plugging in one result after the other and I get the answer. Can't believe you turned it to be sooooooooo easy. :)

You're great guys, thanks! :D