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?