# Homework Help: How to find angle whit system of equations

1. Aug 21, 2012

### felipenavarro

1. The problem statement, all variables and given/known data
i kick a ball with an initial velocity of 18m/s. there is a wall at 8m that has a height of 1.5 meters. whith what angle should i kick the ball so that it bearly passes the wall?

2. Relevant equations
first equation: hf= 1/2gt^2+vit+hi
second equation: V=Δx/t

3. The attempt at a solution
so what i am trying to do is a system of equations isolating t in the second equation and placing it in the first equation. in velosity in the y direction i use 18sin(θ) and for v in the x i use 18cos(θ).
Also as final heigth i am using 1.6m(just a little higher than the wall) and as xfinal i am using 8.1m(also bearly passing the wall)

the problem is i cant soleve the equation. is this the right way to do it?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 21, 2012

### CAF123

So you have $$s_y = v_{oy}t -\frac{1}{2}gt^2,$$ and you have values for $s_y$ and $v_{oy}.$
If you know the wall is 8m away, and you know $v_{ox},$ then by rearranging $s_x = v_{ox}t$ for $t$ and subbing into the above eqn, you should be able to solve for $θ.$
It is just a case of rearranging the final eqn for $θ.$ Is this the part you are having difficulty with?

3. Aug 21, 2012

### felipenavarro

sorry what is sy(changes in y?) , and yes that is where i am having trouble. i end up with cosines and sines and cant unite them or cancel them or anithyng

4. Aug 22, 2012

### ehild

It is the right way in principle but "just a little higher" can mean anything. Why not 1.55 m? or 1.51 m? So choose y=1.5 m at x=8 m.
After eliminating t, you get an equation which contains both the tangent and the cosine of theta. (y=tan(θ)x-(g/2)x2/(182cos2(θ)). You certainly know that cos2(θ)=1/(1+tan2(θ)). Use that relation and you get a quadratic equation for tan(θ).

ehild

Last edited: Aug 22, 2012
5. Aug 22, 2012

### CAF123

$s_y$ is the vertical displacement of the object. I implicitly assumed $s_o = 0$ in the equation $$s_y - s_o = v_{oy}t -\frac{1}{2}gt^2$$

6. Aug 22, 2012

### felipenavarro

now i got it, great explenation! thanks

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