Calculating Optimum Angle for Basketball Throw

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Homework Help Overview

The discussion revolves around calculating the optimum angle for a basketball throw, given the player's height, distance from the basket, and the speed of the throw. The problem involves concepts from projectile motion, specifically relating to the vertical and horizontal components of motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of equations for maximum height and range but question their applicability due to differing heights of the player and the basket. There are discussions about resolving the problem into horizontal and vertical components, and participants suggest creating a new equation that accounts for both components.

Discussion Status

Participants are actively engaging with the problem, offering hints and guidance on how to approach the calculations. There is a recognition of the need to separate the x and y components of motion, and some participants express confusion about specific steps in the calculations, indicating an ongoing exploration of the problem.

Contextual Notes

There are indications of confusion regarding the application of equations and the need for clarity on the definitions of variables. Participants also note the importance of considering the height difference between the player and the basket in their calculations.

  • #31
hayowazzup said:
1 - 5.71059= -4.71059
so the angle is below the horizon? how?

eh? :confused:

cos2θ - 10 sin2θ = - 9800/121 - 1 = -99921/121,

so using tan ψ = 10, and therefore cosψ = 1/√101:

cos2θ - tanψ.sin2θ = -99921/121,

so cosψ.cos2θ - sinψ.sin2θ = -99921/121√101,

so … ? :smile:
 
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  • #32
i thought tan φ = A/B ?
 
  • #33
cosψ.cos2θ - sinψ.sin2θ = -99921/121√101,
cos (ψ+θ) = -99921/121√101,
ψ+θ = -99921/121√101,
θ = -99921/121√101 - ψ

but where the 101 came from?
and i dun get tan ψ = 10 = cosψ = 1/√101
and - 9800/121 - 1 = -99921/121?

i think i m totally lost haha, do you think there's any other easier way to solve this problem?
 
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  • #34
hayowazzup said:
cosψ.cos2θ - sinψ.sin2θ = -99921/121√101,
cos (ψ+θ) = -99921/121√101,
ψ+θ = -99921/121√101,
θ = -99921/121√101 - ψ

erm … it's (ψ+2θ)! :smile:

and however did you get from line 2 to line 3? :rolleyes:
but where the 101 came from?

cosψ = √101, sinψ = 10/√101, tanψ = 10. :smile:
and - 9800/121 - 1 = -99921/121?

erm … nooooooooo …
i think i m totally lost haha, do you think there's any other easier way to solve this problem?

Very possibly … as I said at the start:
tiny-tim said:
The way I'd solve this … and I suspect it's not the quickest :rolleyes: … is to rewrite it in the form A.cos2θ + B.sin2θ = C , then define tan φ = A/B, and solve for (θ - φ). :smile:
 
  • #35
cosψ.cos2θ - sinψ.sin2θ = -99921/121√101,
cos (ψ+2θ) = -99921/121√101,
ψ+θ = cos-1 (-99921/121√101)
θ = cos-1( -99921/121√101 )- 5.71059

but is cos-1( -99921/121√101 ) possible?
 
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  • #36
hayowazzup said:
cosψ.cos2θ - sinψ.sin2θ = -99921/121√101,
cos (ψ+2θ) = -99921/121√101,
ψ+θ = cos-1 (-99921/121√101)
θ = cos-1( -99921/121√101 )- 5.71059

No, it's ψ+ = cos-1 (-99921/121√101),
so = cos-1( -99921/121√101 )- tan-1(10) …
but is cos-1( -99921/121√101 ) possible?
cos is negative for angles > π/2. :smile:

(oh, and 99921 is still wrong, and tan-1(10) isn't 5.71059. :smile:)
 
  • #37
ψ+2θ = cos-1 ((-9800/121-1)√101),
2θ = cos-1((-9800/121-1)√101 )- tan-1(10)
θ = [cos-1((-9800/121-1)√101 )- tan-1(10) ] /2 = -32.6226 ??
 
  • #38
hayowazzup said:
ψ+2θ = cos-1 ((-9800/121-1)√101),
2θ = cos-1((-9800/121-1)√101 )- tan-1(10)
θ = [cos-1((-9800/121-1)√101 )- tan-1(10) ] /2 = -32.6226 ??

hmm … I thought you'd checked up on me? You missed …
tiny-tim said:
1 = 11sinθ (10/11cosθ) - 4.9 (100/121cos²θ)

so 1 = 10sinθ/cosθ - (4900/121)cos²θ

The 4900 should have been 490. :blushing:

So it's:
ψ+2θ = cos-1 ((-980/121-1)√101) … :smile:
 
  • #39
umm
ψ+2θ = cos-1 ((-980/121-1)√101)
θ = -15.6577
y= Vyi *t - (1/2)gt^2 , y= 1.5648 ?
 
  • #40
hayowazzup said:
umm
ψ+2θ = cos-1 ((-980/121-1)√101)
θ = -15.6577
y= Vyi *t - (1/2)gt^2 , y= 1.5648 ?

I'm lost. :confused:

You must show your complete working, or I can't check it! :smile:
 
  • #41
ψ+2θ = cos-1 ((-980/121-1)√101)
θ= [cos-1 ((-980/121-1)√101) - tan-1 (10) ] /2
θ = -15.6577
y= Vyi *t - (1/2)gt^2 ,
y= 11sin(-15.6577)*(10/11cos(-15.6577)) - (1/2)(-9.8)(10/11cos(-15.6577)) ^2
y= 1.5648
 
  • #42
How did you get from θ= [cos-1 ((-980/121-1)√101) - tan-1 (10) ] /2
to θ = -15.6577?
 
  • #43
θ= [cos-1 ((-980/121-1)√101) - tan-1 (10) ] /2
θ= [cos-1 ((-8.099-1)√101) - tan-1 (10) ] /2
θ= [cos-1 ((-9.099)√101) - tan-1 (10) ] /2
θ= [cos-1 (0.6021) - tan-1 (10) ] /2
θ= [52.9739 - 84.289 ] /2
θ= -31.3154 /2
θ = -15.6577
 
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  • #44
how did you get 0.6021? :confused:

(and shouldn't the √101 be on the bottom of the fraction?)
 
  • #45
oh.. i thought 0.6021√101 is 0.6021 to the power of √101
so
θ= [cos-1 ((-980/121-1) / √101) - tan-1 (10) ] /2
θ= [cos-1 ((-8.099-1) / √101) - tan-1 (10) ] /2
θ= [cos-1 ((-9.099) / √101) - tan-1 (10) ] /2
θ= [cos-1 (-0.9054) - tan-1 (10) ] /2
θ= [154.877-84.289 ] /2
θ= 70.588 /2
θ = 35.294


y= Vyi *t - (1/2)gt^2 ,
y= 11sin(35.294)*(10/11cos(35.294)) - (1/2)(9.8)(10/11cos(35.294)) ^2
y= 0.99986
cool, but shouldn't the graivty be negative?
 
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  • #46
hayowazzup said:
oh.. i thought 0.6021√101 is 0.6021 to the power of √101
so
θ= [cos-1 ((-980/121-1) / √101) - tan-1 (10) ] /2
θ= [cos-1 ((-8.099-1) / √101) - tan-1 (10) ] /2
θ= [cos-1 ((-9.099) / √101) - tan-1 (10) ] /2
θ= [cos-1 (-0.9054) - tan-1 (10) ] /2
θ= [154.877-84.289 ] /2
θ= 70.588 /2
θ = 35.294

HI hayowazzup! :smile:

Yes, that looks fine … except you've missed one of the solutions:

cos-1 (-0.9054) can be either 154.877º or 205.123º. :smile:
cool, but shouldn't the graivty be negative?

mmm … not so cool … 'cos this is the solution for it going through the hoop upwards … is that "two points", or "zero points"? :cool:

And not following you about the gravity … the g is minus. :confused:
 
  • #47
tiny-tim said:
HI hayowazzup! :smile:

Yes, that looks fine … except you've missed one of the solutions:

cos-1 (-0.9054) can be either 154.877º or 205.123º. :smile:

oh yep

tiny-tim said:
mmm … not so cool … 'cos this is the solution for it going through the hoop upwards … is that "two points", or "zero points"? :cool:

lol whys that? due to positive acceleration?

tiny-tim said:
And not following you about the gravity … the g is minus. :confused:

i thought acceleration is negative when throwing the ball upwards
im quite confused with that.
 
  • #48
hayowazzup said:
i thought acceleration is negative when throwing the ball upwards
im quite confused with that.

Good morning, hayowazzup! :smile:

Yes, you're right … acceleration is negative (upwards), so the equation needs a minus in front of the g … which it has. :smile:

Why is that worrying you? :confused:
 
  • #49
but if i change it to negative, "y" will be 13.1576
 
  • #50
hayowazzup said:
but if i change it to negative, "y" will be 13.1576

I must be missing the point …

are you using y= Vyi*t - (1/2)gt^2 ?

because the g there is negative.

What are you changing to negative? :confused:
 
  • #51
k i get it now :D thanks
 

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