# Projectile motion

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?

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Staff Emeritus
Gold Member
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?

Homework Helper
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

Homework Helper
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??

Homework Helper
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?

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Homework Helper
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

Homework Helper
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

Homework Helper
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)

Staff Emeritus
Gold Member
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 ?

Staff Emeritus
Gold Member
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)

Homework Helper
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°

Homework Helper
… 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°

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Homework Helper
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

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Homework Helper
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

Homework Helper
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! :yuck:

So use φ = +5.71059.

Then θ is … ?

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

Homework Helper
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?

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Homework Helper
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?

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