What are the two angles of a successful field goal kick?

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Homework Help Overview

The problem involves a placekicker attempting to kick a field goal from a distance of 26.5 meters with an initial velocity of 21.5 m/s. The goal is to determine the two angles at which the ball can clear a crossbar that is 2.81 meters high.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss separating the initial velocity into x and y components and using kinematic equations to find time. There is a focus on substituting values into equations and simplifying expressions involving trigonometric functions.

Discussion Status

Participants have engaged in a back-and-forth exploration of the problem, with some providing guidance on the use of trigonometric identities and algebraic manipulation. There are indications of confusion regarding calculations, and some participants are checking their work against the original equations.

Contextual Notes

There are mentions of potential arithmetic errors and the need to verify calculations against the kinematic equations provided in the problem statement. Participants are working under the constraints of homework rules that require them to explore the problem without providing direct solutions.

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Homework Statement


A placekicker is about to kick a field goal. The ball is 26.5 m from the goalpost. The ball is kicked with an initial velocity of 21.5 m/s at an angle θ above the grojund. Between what two angles, θ1 and θ2, will the ball clear the 2.81-m-high crossbar? Give your answers as (a) the smaller angle and (b) the larger angle.(Hint: The following trigonometric identities may be useful: sec θ=1/(cosθ) and sec2θ=1+tan2θ


Homework Equations


x=26.5m
y=2.81m
v0=21.5m/s
ax=0 m/s
ay=-9.8 m/s

d=v0t+1/2at2


The Attempt at a Solution


Okay I'm stuck. I believe I should separate the initial velocity into an x and y component. Possibly 26.5cosθ=the x component and 26.5sinθ=the y component.
Then maybe plug in those values into the equation to find time? Can someone tell me if I'm anywhere near the right track? The farther along I go the bigger mess of trig functions I get
Please
 
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StephanieF19 said:

The Attempt at a Solution


Okay I'm stuck. I believe I should separate the initial velocity into an x and y component. Possibly 26.5cosθ=the x component and 26.5sinθ=the y component.
Yes, exactly.

Then maybe plug in those values into the equation to find time?
I want to say yes, but it depends which values and which equation, specifically, you are talking about.
 
So...
26.5=(21.5cosθ)t
and
2.81=(21.5sinθ)t+1/2(-9.8)t^2

is that the right next step? Solve the top equation for t and plug that t into the second equation to be able to solve for θ?
 
Okay that sound good in theory :). But my trig isn't that good so i get a little confused.
So far I have
26.5/(21.5cosθ)=t
Plugged it into get:
2.81=(21.5sinθ)(26.5/21.5cosθ)+(-4.9)(26.5/21.5cosθ)^2
Can I get some hint at what I should do next to be able to get rid of all the trig functions
 
Sure!

One term contains sinθ/cosθ, which is equal to ___θ?

The other term has (1/cosθ)^2 ... use the two trig identity hints from the problem statement to change that.
 
Okay. Using that this is what I'm getting:
2.81=26.5tanθ+(-6.04*(1+tan2θ))
So I should distribute?
2.81=26.5tanθ-6.04-6.04tan2θ
Put everything to one side?
0=-6.04tan2θ+26.5tanθ-8.85
What should I do now?
 
Well, if we let x = tanθ, the equation looks like this:

-6.04x^2 + 26.5x - 8.85 = 0​

Can you solve that for x?
 
Oh the quadratic.
Ok so...
-26.5+√702.25-(4*-6.04*-8.85)
-------
2*-6.04
and the same but with a minus
-26.5-√702.25-(4*-6.04*-8.85)
-------
2*-6.04
so x=0.36 and 4.02
Then I plug it back in right? So I get:
tan-10.36=θ and tan-14.02=θ
So θ=19.8 and 76.03?
 
  • #10
StephanieF19 said:
Oh the quadratic.
Ok so...
-26.5+√702.25-(4*-6.04*-8.85)
-------
2*-6.04
and the same but with a minus
-26.5-√702.25-(4*-6.04*-8.85)
-------
2*-6.04
so x=0.36 and 4.02
Then I plug it back in right? So I get:
tan-10.36=θ and tan-14.02=θ
So θ=19.8 and 76.03?

That should work out, but I'm seeing something wrong when I check these angles against these equations:

StephanieF19 said:
So...
26.5=(21.5cosθ)t
and
2.81=(21.5sinθ)t+1/2(-9.8)t^2

Solve the top equation for t and plug that t into the second equation to be able to solve for θ

Okay, so using the 19.8 degrees solution, I solve the 1st equation for t and get 1.31 s.
Then using 19.8 degress and 1.31s in the 2nd equation, I get 1.13 instead of 2.81. So there is a mistake somewhere ...

2.81=(21.5sinθ)(26.5/21.5cosθ)+(-4.9)(26.5/21.5cosθ)^2
.
.
.
2.81=26.5tanθ+(-6.04*(1+tan2θ))
I get something different than the 6.04 number ... I would double-check how you got that.

You have the right method to work this through, it's just a matter of correcting the arithmetic/algebra error.

I'm logging off the computer for tonight. When you work this through and get the angles, remember to check each angle in the equations:
StephanieF19 said:
So...
26.5=(21.5cosθ)t
and
2.81=(21.5sinθ)t+1/2(-9.8)t^2

Solve the top equation for t and plug that t into the second equation to be able to solve for θ

Good luck!
 
  • #11
Okay I got it this time for sure :). Your right my 6.04 answer should have been 7.44 just a miscalculation. Once I worked it through I got 23.8 and 72.2 and those work in the equations :)
Thank you for your help
 

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