# Classic Football Kick Problem

• Loppyfoot
In summary, a football kicker can give the ball an initial speed of 27 m/s. He is 54 m from the goalpost which has a crossbar 3.37 m high. He can kick the football the smallest angle of elevation that he can and score a field goal, or the largest angle of elevation that he can and score a field goal.

## Homework Statement

A football kicker can give the ball an initial speed of 27 m/s. He is 54 m from the goalpost which has a crossbar 3.37 m high.
(a) What is the largest angle of elevation that he can kick the football and score a field goal?

(b) What is the smaller angle of elevation that he can kick the football and score a field goal?

## Homework Equations

rf=r0 + vdeltat
rf=ri+vot+1/2at^2

## The Attempt at a Solution

I'm confused about where to begin this problem ,considering that I don't have time.
Would I do something like:
rf-ri= 27sintheta, but then I would need the change in time, which is not given.

Any thoughts on how to solve this problem?

Next time posting a new topic please include name of topic of question in title, like for this one its kinematics.

For smallest angle you can see that range = 54. so just simply put them in the formula of range. pleas tell me of you don't know.

an for largest angle, when y coordinate is 3.37 during flight, x coordinate should be 54.
Hope it helps.

hmm. I'm still confused.

Would I do

54= 27sin(theta)*deltat + .5(-9.8)deltat

and then solve for theta?

use these eqn

Range = (u2sin2θ)/g
Max height = (u2sin2θ)/2g
Time of flight = (2u sinθ)/g

also at any point,

y = xtanθ - gx2/(2u2 cos2θ)

u stands for velocity, correct?

and how do I differentiate between the largeer and smaller angles?

Last edited:
u is initial velocity
Loppyfoot said:
and how do I differentiate between the largeer and smaller angles?
and use the forth eqn i gave you, for max theta when y is 3.37 .. x must be?

my answers don't seem to be working.

I get for the largest angle possible:

54= 27^2sin(theta)^2/ (9.8) = .8520=sintheta= 58 degrees. would that be correct for the larger angle?

Loppyfoot said:
54= 27sin(theta)*deltat + .5(-9.8)deltat

This is a wrong eqn, when you used d=54 ... i.e. x distance, a=0 as there is no acc. in x axin

2 eqn you might get are:

54 = ucosθ Δt
and
3.37 = usinθΔt + 0.5a(Δt)2

Loppyfoot said:
my answers don't seem to be working.

I get for the largest angle possible:

54= 27^2sin(theta)^2/ (9.8) = .8520=sintheta= 58 degrees. would that be correct for the larger angle?

its sin2θ not sin2θ

and sin2θ and sinθ2 are different

cupid.callin said:
This is a wrong eqn, when you used d=54 ... i.e. x distance, a=0 as there is no acc. in x axin

2 eqn you might get are:

54 = ucosθ Δt
and
3.37 = usinθΔt + 0.5a(Δt)2

Because the problem does not supply deltat, where should that number come from?

54m/27m/s= 2s?

this way you would find smaller angle ... just imagine yourself to be the kicker,
would higher angle be when ball just reaches the post or when it strikes the top of post?

Now tell me how to find the higher angle?

cupid.callin said:
54 = ucosθ Δt
and
3.37 = usinθΔt + 0.5a(Δt)2

you have these two eqn's
and θ, Δt are are unknonw

can you work out a way to find any1 of those?

the higher angle would be when the ball strikes the top of the post.

I attempted to solve for deltat, but when I plugged delta t from the 2nd equation into the first equation i get a long equation with costheta- costhetasintheta.

I'm confused

yes ...

now just consider the ball during its motion ... the point where x becomes 54 apply the condition that y=3.37

use my eqn or general method and you can solve for θ

I don't see what you're getting at. I know there are these two equations, I need to solve a variable deltat to get one variable left. After I try to do that step, I get a ridiculous equation that I need to solve for theta.

Is there any other process of doing this problem?

do it the other way ...

use eqn one to find Δt in terms of θ
substitute that in eqn 2
and solve.

always substitute smaller eqn into larger one

wait i'll try to solve

Ok i'll, as well.

hmm. what did you come out with?

i didnt thought of it but seriously this one is damn irritating.

cant get sinθ out so as to solve cosθ

Are there any sin and cosine laws that would help, like the square of sin added to the square of cosine equalling one?

well there is ...

sinθ = sqrt(1-cos2θ)

You could square both sides, and then we can work it out from there?

well if you square, you can only substitute for sin2θ or cos2θ

well still end up getting something involving sin2θ and sinθ and cosθ
or cos2θ and cosθ and sinθ

wish i had attended my trigo classes

You have Vi = 27, Vix = 27cos[itex]\theta[/tex] and Viy = 27sin[itex]\theta[/tex].

But keeping all your equations in terms of Vix will result in a quadratic in the form of

$$a {V_{ix}}^4 + b {V_{ix}}^2 + c = 0$$

Once you solve for Vix (which will have 2 answers), you can determine the 2 angles.

Say you had a similar problem but you didn't know the initial velocity. Instead you know that you kick the football at 53 degrees, how do you go about finding out what your minimum initial velocity has to be to clear a 10 foot post 36 feet away?