Expert: Scott A Wilson - 4/15/2009

Question
Please help me solve and understand how to solve this Trigonomety Projectile Motion Word Problem. Thanks so much in advance for any help. I have a good understanding of radians, degrees, the relationships of the trig functions, solving identities problems, solving trig equations, graphing the trig functions, etc, but this confuses me. I had a question just like it on my first test and I only got it partially right, only because it was partially the same as a question in my book and I remembered the answers, but I never understood how to solve it.

A projectile fired into the first quadrant from the origin of the coordinate system (0,0) will pass through the point (x,y) at time t according to the relationship cot(theta)=(2x / 2y + gt^2) where theta is the angle of elevation of the launcher and g = the acceleration due to gravity = 32.2 feet/second^2. An artilleryman is firing at an enemy bunker 2425 feet up the side of a hill that is 6050 feet away. He fires a round, and exactly 2.11 seconds later he scores a direct hit.

(a) What angle of elevation did he use?

(b) If the angle of elevation is also given by sec(theta) = v0*t/x, where v0 is the muzzle velocity of the weapon; find the muzzle velocity of the piece he used.

Answer
On the ctn(Θ) = 2x/2y + gt², I don't understand where it comes from.
I also don't know if it is really suppose to be 2x/(2y+gt²) or what.

On (b), I'll ignore sec(Θ) = v0*t/x.
I'm not sure for what that is supposed to be used.
=====================================================================
Since I know how gravity works, I'll do it that way.

(a) Let v = velocity, vx = horizontal velocity, and
vy = vertical velocity.

It can be said that vx is the distance travelled over the time it takes to get there, that is, vx = 6050/2.11.
That is, vx = 2867.298578, which I'll say is 2867,
but save the other for later reference in the spreadsheet as vx.

From physics, the equation for height has g, vy, and the initial height.  The result is the ending height.  The equation is
-gt²/2 + vy*t + 0 = 2452.  That means that vy = (2452+gt²/2)/t.

Since we know t = 2.11 and g = 32.2, put those in to find vy.
vy = (2452 + 32.2*2.11²/2)/2.11.  
Putting that in Excel gives vy = 1196.056308,
which we can round to vy = 1196,
saving the old in the spreadsheet for later reference as vy.

Now if we take those values of vx and vy that I saved, we can find Θ.
The variable Θ is the angle of elevation of the original shot.
I know that ctn(Θ) = dx/dy (which are still in my spreadsheet) = 23°.
It's really 22.6428225°, but I rounded to 23°.  We seem to have 3 digits accuracy, so I could also say that Θ = 22.6°.

(b) The original angle of fire is given by tan(Θ) = vy0/vx0, where
vy0 is the original upward velocity and
vx0 is the original horizontal velocity.

This would mean Θ = atan(vy0/vx0) = 0.395191805 radians,
which is 22.6428225°.  We'll round it to 22.6°.