Finding launch angles for a field goal kick at 42.2 m and 3.75 m height

[chaotixmonjuish]

The kicker on a football team can give the ball an initial speed of 26.2 m/s. If he is to score a field goal from a point 42.2 m in front of goalposts whose horizontal bar is 3.75 m above the ground, what is the minimum and maximum angle above the horizontal he must kick the ball?

I figured out this much

ax=0
ay= -9.8
vx= sin(theta)*26.2
vy= cos(theta)*26.2
x= sin(theta)*26.2x
y= cos(theta)*26.2x-4.9x^2

and I think

3.75= cos(theta)*26.2x-4.9x^2

My first attempt at an answer yielded 5.06 and 26.something. I tried using arctan(3.75/42.2) to get an angle. I'm not really sure how to approach this problem since theta is missing.

 

[learningphysics]

By 'x', do you mean time? I'm a little confused by 'x' when it is used for multiplication along with as a variable...

 

[chaotixmonjuish]

yes, x for time

 

[learningphysics]

You have the vx, vy mixed...

It should be:
vx= cos(theta)*26.2
vy= sin(theta)*26.2

So the two equation become (I'm going to switch to t becomes we have another x already ):

42.2= cos(theta)*26.2t
3.75= sin(theta)*26.2t-4.9t^2

solve these to get theta.

Then you also should solve:
42.2= cos(theta)*26.2t
0= sin(theta)*26.2t-4.9t^2

For this case, the second equation simplifies to:
0= sin(theta)*26.2-4.9t

by dividing both sides by t.