What Angle Should a Football Kicker Use for Maximum Hang Time on a 65 Yard Kick?

In summary: I am going to bed right after I understand your post =DThanks guys.In summary, the conversation discusses a football kicker's desire to kick the ball from his own 34 yard line to the 1 yard line of the opposing team, which is a 65 yard kick. The kicker can kick the ball at 27 yards/sec. The team wants to give their own teammates as much time as possible to reach the ball, so the conversation involves finding the angle at which the kicker should kick the ball in order to achieve maximum hang time. The conversation also discusses the use of trigonometry and equations to solve the problem.
  • #36
ok when i multiply both sides by Vo*Cosθ, I got...
65= T = 2(Vo*sinθ)(27cosθ)/g
 
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  • #37
ok i got the
65 = 2*Vo*Sinθ*Cosθ
but how did you get the...

65 = 2*Vo*Sinθ*Cosθ = Vo*Sin2θ/g
 
  • #38
ohh ok I got here now!

65 = 2*Vo2*Sinθ*Cosθ/g = Vo2*Sin2θ/g

since
Vo2*Sin2θ/g = 65, how would I isolate the θ?
 
Last edited:
  • #39
ok what I did now is,
65 = 272sin2θ / g
(65*(-9.8)) / 272 = sin2θ

how do I isolate the θ now?
it's the final question then I think that would be the answer afterwards =P
 
  • #40
hmm...I got -30.46o which means the person is shooting the ball into the ground and the ball is going at super high speeds which makes it eat the dirt and make a parabolic tunnel in the ground to the other side? loool wth?
 
  • #41
twenty5 said:
hmm...I got -30.46o which means the person is shooting the ball into the ground and the ball is going at super high speeds which makes it eat the dirt and make a parabolic tunnel in the ground to the other side? loool wth?

You got the right angle if you are supposed to treat g as 9.8 yds/s2.

But the wrong sign.

Except you need to recognize that there is another angle for which there is a solution. Consider that Sin(θ) = Sin(180-θ)

What is 1/2 of that angle?

And won't that angle be greater with respect to the horizon ... meaning longer time of flight?
 
  • #42
LowlyPion said:
You got the right angle.

But the wrong sign.

Except you need to recognize that there is another angle for which there is a solution. Consider that Sin(θ) = Sin(180-θ)

What is 1/2 of that angle?

And won't that angle be greater with respect to the horizon ... meaning longer time of flight?

sin(θ) = sin(180 - 30.46)?

or do you mean, that θ is in the top left quadrant O_O... anyhow... sin(149.54) and divide that angle by 2 would give me.. 74o or so O_O so which one would be correct?
 
  • #43
twenty5 said:
sin(θ) = sin(180 - 30.46)?

or do you mean, that θ is in the top left quadrant O_O... anyhow... sin(149.54) and divide that angle by 2 would give me.. 74o or so O_O so which one would be correct?

No.

What was the 2θ angle you found?

180 - 2θ = 90 - θ.
 
  • #44
LowlyPion said:
No.

What was the 2θ angle you found?

180 - 2θ = 90 - θ.

I found that θ = -30.46o

from...

-0.874 = sin2θ

sin-1 (-0.874) = 2θ
 

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