Can't remember how I got these answers

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 angle above the horizontal he must kick the ball?

I was suppose to calculate the minimum and maximum angles.

I got 24.3 and 70.8.

I set up this equation:
xf=xi+vi*cos(theta)t
yf=yi+vi*sin(theta)t+4.9t^2

if you solve x time, and then sub it in, it comes out to

yf=yi+tan(theta)x+(4.9x^2/v^2)tan^2(theta)

3.75=0+tan(theta)42.2+(4.9(42.2)^2/26.2^2)tan^2(theta)

I'm not sure where I'm making the error.

I think I did this right, we were told to use the quadratic formula and treat the tan as variables. I'm not really sure what I did after that however.

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 angle above the horizontal he must kick the ball?

I was suppose to calculate the minimum and maximum angles.

I got 24.3 and 70.8.

I set up this equation:
xf=xi+vi*cos(theta)t
yf=yi+vi*sin(theta)t+4.9t^2

if you solve x time, and then sub it in, it comes out to

yf=yi+tan(theta)x+(4.9x^2/v^2)tan^2(theta)

3.75=0+tan(theta)42.2+(4.9(42.2)^2/26.2^2)tan^2(theta)

I'm not sure where I'm making the error.

I think I did this right, we were told to use the quadratic formula and treat the tan as variables. I'm not really sure what I did after that however.

You have made a mistake in deriving the final formula. Try deriving the formula again.

I re-derived it and got the same thing.

I'm not really finding an error, it must be something I missed at the very beginning.

Would you be referring to the 4.9's position within the parantheses. I didn't have it like that in my calculations.

I re-derived it and got the same thing.

Ok let me tell you that the equation is

y = $$\tan\theta$$x - $$\frac{gx^2}{2v^2\cos^2\theta}$$

yes, but doesn't 1/cos^2=1+tan^2

ahhh i forgot to sub in 1+tan^2

yes, but doesn't 1/cos^2=1+tan^2

ahhh i forgot to sub in 1+tan^2

Of course it is.But you didnt substitute it. Ok,so got the solution.

Since the hint was to plug that back into the quadratic formula. I'm just not sure what the C value would be.

I'm still getting an answer that's way off

I'm still getting an answer that's way off

Cmon its taking too long.Its now just a quadratic equation.You can solve it now with much ease.But if you are not able to get then i am here.

I have got the answer but i cannot give it to you as it will break the forum rules.