Calculating Optimum Angle for Basketball Throw

[hayowazzup]

Homework Statement

A bastketball player is 2m tall and stands 10m from the basket.The height of the basket is 3m and the player throws the ball with speed 11m/s. At what angle with respect to the horizontal should he throw the ball so that is goes through to basket?

Homework Equations

g=gravity, h= maximum height vi= initial velocity x=angle R= maximum range
h=vi^2 * (sin(x))^2 / 2g
R= vi^2 * sin 2x /g

The Attempt at a Solution

Here is what've done:

h=vi^2 * (sin(x))^2 / 2g
x= Inverse SIN (SQRT ( h*2g / vi^2 ))
height = 1m
x= Inverse Sin(SQRT (1*2(9.8)/ 11^2)) = 23.73 degree

R= vi^2 * sin 2x /g
x= (Inverse SIN(R*g / vi^2)) / 2
R= 10m
x= (Inverse SIN(10*9.8 / 11^2)) / 2 = 27.04degree

both angles aint the same,
is it because the height of the basket is different to the player's height?
do i need to put "h=vi^2 * (sin(x))^2 / 2g " and
"R= vi^2 * sin 2x /g " into one eqn in order to get the angle?

 

[Kurdt]

You need to resolve the problem into horizontal and vertical components. What is the equation describing the horizontal motion of the ball? what is the equation tha describes the vertical motion of the ball?

 

[tiny-tim]

A bastketball player is 2m tall and stands 10m from the basket.The height of the basket is 3m and the player throws the ball with speed 11m/s. At what angle with respect to the horizontal should he throw the ball so that is goes through to basket?

Homework Equations

g=gravity, h= maximum height vi= initial velocity x=angle R= maximum range
h=vi^2 * (sin(x))^2 / 2g
R= vi^2 * sin 2x /g
…
is it because the height of the basket is different to the player's height?

Hi hayowazzup!

Doom … despair … calamity …

Those equations won't help.

You don't need the maximum height, and you don't need the range (which, as you say, is only for heights which are the same).

You'll have to start from scratch, and work out your own equation.

Hint: deal with the x and y components separately … oh, and don't call the angle x as well, that'll only confuse you … have a theta …

 

[hayowazzup]

thnx
so,
cos θ=Vxi / Vi => Vxi=cos θ * Vi
sin θ=Vyi/ Vi => Vyi=sin θ * Vi

 

[tiny-tim]

thnx
so,
cos θ=Vxi / Vi => Vxi=cos θ * Vi
sin θ=Vyi/ Vi => Vyi=sin θ * Vi

That's right!

So find out from Vxi how long it takes to go 10m horizontally.

Then concentrate on Vyi, and work out a formula to tell you what its height is at that time.

(And remember that the height has to be +1m for the second time … the ball has to go through from above!).

 

[hayowazzup]

∆x = Vx0 * t
t= ∆x / Vx0
t= 10m / Vx0
but how do i find out the initial horizontal velocity??

 

[tiny-tim]

t= 10m / Vx0

That's right!

So t = 10/11cosθ.

Vy0 = 11sinθ, so what is the height at t = 11cosθ?

 

[hayowazzup]

do you mean the Vyf?
0m?

 

[tiny-tim]

do you mean the Vyf? is it 0m?

I'm confused.

Vyf is a velocity.

Hint: ay = -9.8. Vy0 = 11sinθ. So what is the formula for y (as a function of t)?

 

[hayowazzup]

o i see,
∆y =Vy0 * t - (1/2)g t^2
∆y = 11sinθ * t - (1/2) *-9.8 * t^2

 

[tiny-tim]

o i see,
∆y =Vy0 * t - (1/2)g t^2
∆y = 11sinθ * t - (1/2) *-9.8 * t^2

That's right!

So the height at t = 11cosθ is … ?

 

[hayowazzup]

∆y = 11sinθ * (11cosθ) - (1/2) *-9.8 * (11cosθ) ^2
∆y = 121 sinθ *cosθ + 592.9 (cosθ)^2

 

[tiny-tim]

o i see,
∆y =Vy0 * t - (1/2)g t^2
∆y = 11sinθ * t - (1/2) *-9.8 * t^2

hmm … I've just noticed there's too many minuses in there.

∆y = 11sinθ * (11cosθ) - (1/2) *-9.8 * (11cosθ) ^2
∆y = 121 sinθ *cosθ + 592.9 (cos²θ)

ok … now put ∆y = 1, and solve using usual trig methods.

 

[hayowazzup]

unfortunately, i don't have a clue about how to solve the equation..
can you teach me how to do it , or do you have any references that I can look at?thnx
I guess the first step is that I need to factorise it
1 = cos θ(121 sinθ + 592.9 (cos θ))
1/cos θ = 121 sinθ + 592.9 (cos θ)

 

[Kurdt]

Something has gone slightly wrong somewhere. There should be 1/cos θ's hanging around.

 

[hayowazzup]

do you mean shoudnt?
should i use the rule sin² + cos²=1
but then how do i get rid of sinθ *cosθ ?

 

[Kurdt]

Nope I mean you worked out the time with the x-component of the motion and that came out as t = 10/11 cos θ.

 

[hayowazzup]

1 = 11sinθ (10/11cos²θ) + 4.9 (10/11cos²θ)

 

[tiny-tim]

Hi hayowazzup!

(btw, it's 1 = 121 sinθ *cosθ - 592.9 (cos²θ), with a minus in the middle)

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]

hi, right so,
sin 2θ= 2sinθcosθ
cos 2θ= 2cos²θ -1

1 = 121 sinθ *cosθ - 592.9 (cos²θ)
0 = 60.5 (2sinθcosθ) - 296.45 * 2(cos²θ) - 1
0 = 60.5 sin 2θ - 296.45 cos 2θ
tan φ= -296.45 / 60.5 = -4.9
φ = tan-1 (-4.9) = -78.46°

 

[tiny-tim]

… oops!

hi, right so,
sin 2θ= 2sinθcosθ
cos 2θ= 2cos²θ -1

1 = 121 sinθ *cosθ - 592.9 (cos²θ)
0 = 60.5 (2sinθcosθ) - 296.45 * 2(cos²θ) - 1
0 = 60.5 sin 2θ - 296.45 cos 2θ
tan φ= -296.45 / 60.5 = -4.9
φ = tan-1 (-4.9) = -78.46°

Hi hayowazzup!

I didn't see posts #17 and 18 when I made my last post.

(or my own misprint in post #7 … thankyou, Kurdt )

So it should be (and you'd better check) …

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

so 1 = 10sinθ/cosθ - (4900/121)cos²θ

and then proceed as before.

(btw your 0 = 60.5 sin 2θ - 296.45 cos 2θ
doesn't follow from 0 = 60.5 (2sinθcosθ) - 296.45 * 2(cos²θ) - 1
you've done it as if it was
0 = 60.5 (2sinθcosθ) - 296.45 * (2(cos²θ) - 1) )

 

[hayowazzup]

1 = 10sinθ/cosθ - (4900/121)(1/cos²θ)
1= (100sin²θ - (4900/121) ) / cos²θ
cos²θ = 100sin²θ - (4900/121)
cos²θ - 100sin²θ = -(4900/121)
k...i think i m stuck :(

 

[hayowazzup]

sin²θ=(1/2)(1-cos 2θ)
cos²θ=(1/2)(1+cos 2θ)

cos²θ - 100sin²θ = -(4900/121)

1/2(1+cos2θ) - 100(1/2)(1-cos2θ) = -(4900/121)
(1+cos2θ) - 100(1-cos2θ) = -(4900/121) * 2
1 + cos2θ -100 + 100cos2θ = -(9800/121)
-99 + 101cos2θ = -(9800/121)
101cos2θ = -(9800/121) + 99
cos2θ = [-(9800/121) + 99] / 101
2θ = cos-1([-(9800/121) + 99] / 101 )
θ = cos-1([-(9800/121) + 99] / 101 ) / 2 = 39.86°

 

[tiny-tim]

1 = 10sinθ/cosθ - (4900/121)(1/cos²θ)
1= (100sin²θ - (4900/121) ) / cos²θ
cos²θ = 100sin²θ - (4900/121)
cos²θ - 100sin²θ = -(4900/121)
k...i think i m stuck :(

hmm … where did your 100 come from?

1 = 10sinθ/cosθ - (4900/121)(1/cos²θ)

so cos²θ = 10sinθcosθ - 4900/121

so 2cos²θ - 1 = 20sinθcosθ - 9800/121 - 1.

so … ?

 

[hayowazzup]

because i was trying to add them together, as i made their denominator the same i square
10sinθ/cosθ = 100sin²θ/cos²θ

 

[hayowazzup]

hmm … where did your 100 come from?

1 = 10sinθ/cosθ - (4900/121)(1/cos²θ)

so cos²θ = 10sinθcosθ - 4900/121

so 2cos²θ - 1 = 20sinθcosθ - 9800/121 - 1.

so … ?

cos 2θ = 10 sin2θ - 9800/121 - 1.
cos 2θ - 10 sin2θ = - 9800/121 - 1

 

[tiny-tim]

because i was trying to add them together, as i made their denominator the same i square
10sinθ/cosθ = 100sin²θ/cos²θ

ah … I wondered where the sin²θ came from!

Nooo … it's 10sinθ/cosθ = 10sinθcosθ/cos²θ, isn't it?

cos 2θ = 10 sin2θ - 9800/121 - 1.
cos 2θ - 10 sin2θ = - 9800/121 - 1

That's right!

Now use the tan method, as before!

(You are checking my figures, aren't you? I've been wrong once already, remember! )

 

[hayowazzup]

yep
tan φ = 1/ -10
φ = tan-1 (-1/10) = -5.71059

 

[tiny-tim]

yep
tan φ = 1/ -10
φ = tan-1 (-1/10) = -5.71059

Yes … except my very strong advice is always to go with positive angles, or you run the grave risk of getting the signs wrong later!

So use φ = +5.71059.

Then θ is … ?

 

[hayowazzup]

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

 

[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?