Can't remember how I got these answers

  • #1
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.
 

Answers and Replies

  • #2
318
0
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.
 
  • #3
I re-derived it and got the same thing.
 
  • #4
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.
 
  • #5
318
0
I re-derived it and got the same thing.

Ok let me tell you that the equation is

y = [tex]\tan\theta[/tex]x - [tex]\frac{gx^2}{2v^2\cos^2\theta}[/tex]
 
  • #6
yes, but doesn't 1/cos^2=1+tan^2

ahhh i forgot to sub in 1+tan^2
 
  • #7
318
0
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.
 
  • #8
Since the hint was to plug that back into the quadratic formula. I'm just not sure what the C value would be.
 
  • #9
I'm still getting an answer that's way off
 
  • #10
318
0
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.
 

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