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

[twenty5]

No. It's 65 yards. It has to cross the 50 to get to the other 1 yard line.

Besides if the problem says it's 65 yards that's how I would roll.

yah sorry I editted, it's 65 yards =P

 

[twenty5]

ok I am trying to figure how this happened at the moment



from :

65/Vo*Cosθ = T = 2*Vo*Sinθ/g


to :

65 = 2*Vo2*Sinθ*Cosθ/g = Vo2*Sin2θ/g

 

[twenty5]

I can't seem to get to where you are at X_X sorrryyy

 

[LowlyPion]

65/Vo*Cosθ = T = 2*Vo*Sinθ/g

Multiply both sides by Vo*Cosθ

 

[twenty5]

Multiply both sides by Vo*Cosθ

both sides including t? or just the very right side?

 

[twenty5]

ok when i multiply both sides by Vo*Cosθ, I got...
65= T = 2(Vo*sinθ)(27cosθ)/g

 

[twenty5]

ok i got the
65 = 2*Vo*Sinθ*Cosθ
but how did you get the...

65 = 2*Vo*Sinθ*Cosθ = Vo*Sin2θ/g

 

[twenty5]

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?

 

[twenty5]

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

 

[twenty5]

hmm...I got -30.46^o 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?

 

[LowlyPion]

hmm...I got -30.46^o 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?

 

[twenty5]

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.. 74^o or so O_O so which one would be correct?

 

[LowlyPion]

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.. 74^o or so O_O so which one would be correct?

No.

What was the 2θ angle you found?

180 - 2θ = 90 - θ.

 

[twenty5]

No.

What was the 2θ angle you found?

180 - 2θ = 90 - θ.

I found that θ = -30.46^o

from...

-0.874 = sin2θ

sin^-1 (-0.874) = 2θ