Finding the Angle to Hit a Window 10m Away and 5m High

[blaziken's_charizard]

1. A boy can throw a stone at 14 m/s. At what angle must he throw it in order to hit a window which is 10m away horizontally and 5m above the ground?
2. I would assume we have to use trig ratios and the kinematics equations.

v = u + a.t
v^2 = u^2 + 2.a.s
s = u.t + 1/2.a.t^2


3. Diagram:

(see attachment or http://img377.imageshack.us/my.php?image=imgwn9.png )I know that time is a constant for both the horizontal and vertical components, so I presume that it would not affect the equations.

My info:
-Vertical motion
(t=0)
s = 5m
u = 14 m/s sin.(theta)
a = +9.8 m/s^2

-Horizontal motion
(t=0)
s = 10m
u = 14 m/s cos.(theta)
a= 0

So how do I find theta? Am I missing any equations? Is my diagram correct? Is u = 14 m/s sin.(theta) and u = 14 m/s cos.(theta) correct? Thanks :)

I'm thinking of using some equation in theta... but what those equations are is beyond me =[

 

[fikus]

you should write how horizontal and vertical distances depend on t, then express t from one equation (horizontal makes it easier), and put it in other.

 

[Mentallic]

There is more manipulation of trigonometric identities involved in this question than actually using formulas.
You have the horizontal velocity vector. Now you want to find how long it will take the projectile to hit the window. This won't be a direct answer, but in terms of \theta.
Use t=\frac{s_x}{u_x}
where,
s_x=horizontal displacement
u_x=horizontal velocity vector

Then from here you can substitute the time into:
s_y=u_yt+\frac{1}{2}a_yt^2
where,
s_y=vertical displacement
u_y=vertical velocity vector
a_y=acceleration due to gravity (make sure this is negative, since it is acting opposite to the velocity vector and the displacement) i.e. -9,8ms^-2

You will now find that you have an equation all in terms of \theta. Now comes the manipulation. You will be surprised to find that there are two possible angles to throw the projectile at

 

[Mentallic]

If you're curious to know, I made some calculations and found for this scenario, but instead being under the influence of stronger gravity (being on another planet); the gravity can only go as high as g=\frac{49(1-\sqrt{5})}{5} \approx -12.11ms^{-2} until the rock would never be able to reach the window. (not taking into account that higher gravity will slow the initial velocity of the throw).
At this gravity, the precise angle to throw the rock at would be tan^{-1}(\frac{\sqrt{5}+1}{2}) \approx 58^o17'

 

[blaziken's_charizard]

There is more manipulation of trigonometric identities involved in this question than actually using formulas.
You have the horizontal velocity vector. Now you want to find how long it will take the projectile to hit the window. This won't be a direct answer, but in terms of \theta.
Use t=\frac{s_x}{u_x}
where,
s_x=horizontal displacement
u_x=horizontal velocity vector

Then from here you can substitute the time into:
s_y=u_yt+\frac{1}{2}a_yt^2
where,
s_y=vertical displacement
u_y=vertical velocity vector
a_y=acceleration due to gravity (make sure this is negative, since it is acting opposite to the velocity vector and the displacement) i.e. -9,8ms^-2

You will now find that you have an equation all in terms of \theta. Now comes the manipulation. You will be surprised to find that there are two possible angles to throw the projectile at

t = s/u
s = 10m
u = 14cos.\theta m/s [horizontal velocity vector]

so, t = 10 / 14cos.\theta

then, s = u.t + 1/2.a.t^2

5 = (14sin\theta * 10/14cos\theta) + 1/2.(-9.8).(10/14cos\theta)(10/14cos\theta)

Please excuse the absence of units- including them would make it so much more confusing.

I will deal with each part of that equation separately:

ut:
(14sin\theta * 10/14cos\theta)
= 140sin\theta / 14cos\theta
Right? I don't know how to continue frome here...

1/2 * a:
1/2.(-9.8)
= -.49

t^2
(10/14cos\theta)(10/14cos\theta)
= 100 / 14cos^2\theta
Don't know how to continue from here either...

So, I have:

5 = (140sin\theta / 14cos\theta) + (-4.9)(100 / 14cos^2\theta)

I guess my problem now is working with the trig identities now...

 

[Mentallic]

ok first we will simplify the equation (just the coefficients at first. i.e. the numbers).

5=14sin\theta(\frac{10}{14cos\theta})+\frac{1}{2}(-9.8)(\frac{10}{14cos\theta})^2

simplified: 5=\frac{10sin\theta}{cos\theta}-4.9(\frac{25}{49cos^2\theta})

Now the trig identities you should be aware of are as follows. If you want to know the proofs for them, just ask

1) tan\theta=\frac{sin\theta}{cos\theta}
thus tan^2\theta=\frac{sin^2\theta}{cos^2\theta}

2) sin^2\theta+cos^2\theta=1

3) sec^2\theta=1+tan^2\theta N.B. sec\theta=\frac{1}{cos\theta}

See if you can apply these formulas in a variety of ways to get an equation all in terms of the same trig ratio. tangent, cosine or sine. Good luck!