Calculating Optimum Angle for Basketball Throw

[tiny-tim]

1 - 5.71059= -4.71059
so the angle is below the horizon? how?

eh?

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 … ?

 

[hayowazzup]

i thought tan φ = A/B ?

 

[hayowazzup]

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?

 

[tiny-tim]

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)!

and however did you get from line 2 to line 3?

but where the 101 came from?

cosψ = √101, sinψ = 10/√101, tanψ = 10.

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:

The way I'd solve this … and I suspect it's not the quickest … is to rewrite it in the form A.cos2θ + B.sin2θ = C , then define tan φ = A/B, and solve for (θ - φ).

 

[hayowazzup]

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?

 

[tiny-tim]

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+2θ = cos-1 (-99921/121√101),
so 2θ = cos-1( -99921/121√101 )- tan-1(10) …

but is cos-1( -99921/121√101 ) possible?

cos is negative for angles > π/2.

(oh, and 99921 is still wrong, and tan-1(10) isn't 5.71059. )

 

[hayowazzup]

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 ??

 

[tiny-tim]

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 …

1 = 11sinθ (10/11cosθ) - 4.9 (100/121cos²θ)

so 1 = 10sinθ/cosθ - (4900/121)cos²θ

The 4900 should have been 490.

So it's:
ψ+2θ = cos-1 ((-980/121-1)√101) …

 

[hayowazzup]

umm
ψ+2θ = cos-1 ((-980/121-1)√101)
θ = -15.6577
y= Vyi *t - (1/2)gt^2 , y= 1.5648 ?

 

[tiny-tim]

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.

You must show your complete working, or I can't check it!

 

[hayowazzup]

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

 

[tiny-tim]

How did you get from θ= [cos-1 ((-980/121-1)√101) - tan-1 (10) ] /2
to θ = -15.6577?

 

[hayowazzup]

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

 

[tiny-tim]

how did you get 0.6021?

(and shouldn't the √101 be on the bottom of the fraction?)

 

[hayowazzup]

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?

 

[tiny-tim]

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!

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

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

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"?

And not following you about the gravity … the g is minus.

 

[hayowazzup]

HI hayowazzup!

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

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

oh yep

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

lol whys that? due to positive acceleration?

And not following you about the gravity … the g is minus.

i thought acceleration is negative when throwing the ball upwards
im quite confused with that.

 

[tiny-tim]

i thought acceleration is negative when throwing the ball upwards
im quite confused with that.

Good morning, hayowazzup!

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

Why is that worrying you?

 

[hayowazzup]

but if i change it to negative, "y" will be 13.1576

 

[tiny-tim]

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?

 

[hayowazzup]

k i get it now :D thanks