Calculating Optimum Angle for Basketball Throw

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The discussion centers on calculating the optimum angle for a basketball throw, given a player height of 2m, a basket height of 3m, and a throw distance of 10m with an initial speed of 11m/s. The equations of motion used include the vertical height equation \( h = \frac{v_i^2 \sin^2(\theta)}{2g} \) and the range equation \( R = \frac{v_i^2 \sin(2\theta)}{g} \). The participants concluded that the angles calculated for height and range differ due to the height difference between the player and the basket, necessitating a combined approach to solve for the angle accurately.

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  • #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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