# Homework Help: Finding launch angle from vertical and horizontal distance

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1. Oct 31, 2015

### vetgirl1990

1. The problem statement, all variables and given/known data
The mountain rescue officers trigger the avalanche by firing a gun at a distant mountain slope.
The muzzle velcoity of the gun is 225m/s. The horizontal and vertical distance from the gun to the slope is
x=1,072m and y=538m respectively. Find the lowest angle between the barrel of the gun and the horizontal plane.

2. Relevant equations
rf = ri + vt + 1/2at2
Dividing each into the horizontal and vertical components:
- Horizontal component: xf = xi + vxt + 1/2axt2
- Vertical component: yf = yi + vyt + 1/2ayt2

Vxi = VicosΘ

Vyi = VisinΘ

3. The attempt at a solution
I started off by manipulating the equations in terms of time (t), since I don't have that either.

In the x direction, the bullet is under constant velocity, so a=0; xf = 1072m; xi = 0
xf = xi + vxt + 1/2axt2
1072 = 225cosΘ t
(1) t = 1072 / 225cosΘ

In the y direction, the bullet is under constant acceleration due to gravity, ay = -g = -9.8m/s2
yf = yi + vyt + 1/2ayt2
528 = 225sinΘ t - 4.9t2
(2) 538 = t(225sinΘ - 4.9t)

Then I substituted equation (1) into (2)...
538 = [1072/225cosΘ] [225sinΘ - 4.9(1072/225cosΘ)
538 = 1072sinΘ/cosΘ - 111.22966/cos2Θ
and this is where the trig substitutions start getting messy. It gets quite long, and I ended up with 1.211249 = sin2Θ, which is impossible.

So either I am approaching the question wrong, or I am not manipulating my trig properly.

Any help or direction would be greatly appreciated.

2. Oct 31, 2015

### Staff: Mentor

That has an actual solution, so your mistake is somewhere in the part you didn't show.

Solving for the cosine is probably a bit easier.

3. Oct 31, 2015

### SteamKing

Staff Emeritus
It's not clear how you went from this:
538 = 1072sinΘ/cosΘ - 111.22966/cos2Θ

to this:
1.211249 = sin2Θ

It would seem that there is a problem with your algebra of trig here which need double checking.

4. Oct 31, 2015

### J Hann

If it helps any the equation for the distance along the plane is
D = 2 * v^2 ^ cos^2 alpha * (tan alpha - tan beta) / (g * cos beta)
where alpha is the angle of the projectile and beta the angle of the plane.
It would seem that the solution for this particular configuration could be found without deriving this formula,
but I admit to never having tried it.
Incidentally, the equation is derived using
y = - g t^2 /2 + v t sin alpha
x = v t cos alpha
and eliminating the time from these equations.
Good Luck and Happy Halloween!

5. Feb 13, 2018

### Capaverde

I've managed to solve it.
If you set everything in terms of tan(theta), then do a substitution, you get a quadratic equation.

Since this hasn't yet been solved I thought it'd be okay to provide my solution, even if this is an old thread.

You got to
538 = 1072sinΘ/cosΘ - 111.22966/((cosΘ)^2)
Which is
538 = 1072tanΘ - 111.23(secΘ)^2
But (secΘ)^2 = (tanΘ)^2 + 1, so
538 = 1072tanΘ - 111.23((tanΘ)^2 + 1)
538 = 1072tanΘ - 111.23(tanΘ)^2 - 111.23
111.23(tanΘ)^2 - 1072tanΘ + 649.23 = 0
Dividing everything by 111.23 we get
(tanΘ)^2 - 9.6377tanΘ + 5.8368 = 0
Substituting tanΘ = u,
u^2 - 9.6377u + 5.8368 = 0
is a quadratic equation with a=1, b=-9.6377, c=5.8368
So,
u = (-b +/- sqrt(b^2 - 4ac))/(2a) = (9.6377 +/- sqrt(92.8853 - 23.3472))/2
u_1 = (9.6377 + sqrt(69.5381))/2 = (9.6377 + 8.3390)/2 = 17.9767/2 = 8.9884
u_2 = (9.6377 - 8.3390)/2 = 1.2987/2 = 0.6494

tanΘ = u
Θ = arctan(u)

There are two solutions

Θ_1 = arctan(u_1) = arctan(8.9884) = 83.6517 degrees = 1.46 radians
Θ_2 = arctan(u_2) = arctan(0.6494) = 32.9997 degrees = 0.5760 radians

Hope I've helped.

6. Feb 13, 2018

### haruspex

Yes and no.
It is far better to solve problems symbolically, only plugging in numbers at the end. It has many advantages.
If the displacements are x and -y then we have
x = vt cos(θ)
y = ½gt2-vt sin(θ)
y = ½gx2sec2(θ)/v2- x tan(θ)
y/x = ½gx(1+tan2(θ))/v2- tan(θ)
etc.