# Homework Help: 2-dimential motion

1. Sep 19, 2009

### StephanieF19

1. The problem statement, all variables and given/known data
A placekicker is about to kick a field goal. The ball is 26.5 m from the goalpost. The ball is kicked with an initial velocity of 21.5 m/s at an angle θ above the grojund. Between what two angles, θ1 and θ2, will the ball clear the 2.81-m-high crossbar? Give your answers as (a) the smaller angle and (b) the larger angle.(Hint: The following trigonometric identities may be useful: sec θ=1/(cosθ) and sec2θ=1+tan2θ

2. Relevant equations
x=26.5m
y=2.81m
v0=21.5m/s
ax=0 m/s
ay=-9.8 m/s

d=v0t+1/2at2

3. The attempt at a solution
Okay I'm stuck. I believe I should separate the initial velocity into an x and y component. Possibly 26.5cosθ=the x component and 26.5sinθ=the y component.
Then maybe plug in those values into the equation to find time? Can someone tell me if I'm anywhere near the right track? The farther along I go the bigger mess of trig functions I get
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 19, 2009

### Redbelly98

Staff Emeritus
Welcome to Physics Forums

Yes, exactly.

I want to say yes, but it depends which values and which equation, specifically, you are talking about.

3. Sep 19, 2009

### StephanieF19

So...
26.5=(21.5cosθ)t
and
2.81=(21.5sinθ)t+1/2(-9.8)t^2

is that the right next step? Solve the top equation for t and plug that t into the second equation to be able to solve for θ?

4. Sep 19, 2009

### Redbelly98

Staff Emeritus
Yes.

5. Sep 19, 2009

### StephanieF19

Okay that sound good in theory :). But my trig isn't that good so i get a little confused.
So far I have
26.5/(21.5cosθ)=t
Plugged it in to get:
2.81=(21.5sinθ)(26.5/21.5cosθ)+(-4.9)(26.5/21.5cosθ)^2
Can I get some hint at what I should do next to be able to get rid of all the trig functions

6. Sep 19, 2009

### Redbelly98

Staff Emeritus
Sure!

One term contains sinθ/cosθ, which is equal to ___θ?

The other term has (1/cosθ)^2 ... use the two trig identity hints from the problem statement to change that.

7. Sep 19, 2009

### StephanieF19

Okay. Using that this is what I'm getting:
2.81=26.5tanθ+(-6.04*(1+tan2θ))
So I should distribute?
2.81=26.5tanθ-6.04-6.04tan2θ
Put everything to one side?
0=-6.04tan2θ+26.5tanθ-8.85
What should I do now?

8. Sep 19, 2009

### Redbelly98

Staff Emeritus
Well, if we let x = tanθ, the equation looks like this:

-6.04x^2 + 26.5x - 8.85 = 0​

Can you solve that for x?

9. Sep 19, 2009

### StephanieF19

Ok so....
-26.5+√702.25-(4*-6.04*-8.85)
-------
2*-6.04
and the same but with a minus
-26.5-√702.25-(4*-6.04*-8.85)
-------
2*-6.04
so x=0.36 and 4.02
Then I plug it back in right? So I get:
tan-10.36=θ and tan-14.02=θ
So θ=19.8 and 76.03?

10. Sep 19, 2009

### Redbelly98

Staff Emeritus
That should work out, but I'm seeing something wrong when I check these angles against these equations:

Okay, so using the 19.8 degrees solution, I solve the 1st equation for t and get 1.31 s.
Then using 19.8 degress and 1.31s in the 2nd equation, I get 1.13 instead of 2.81. So there is a mistake somewhere ...

I get something different than the 6.04 number ... I would double-check how you got that.

You have the right method to work this through, it's just a matter of correcting the arithmetic/algebra error.

I'm logging off the computer for tonight. When you work this through and get the angles, remember to check each angle in the equations:
Good luck!

11. Sep 19, 2009

### StephanieF19

Okay I got it this time for sure :). Your right my 6.04 answer should have been 7.44 just a miscalculation. Once I worked it through I got 23.8 and 72.2 and those work in the equations :)