# Homework Help: Define Angle of Projection with The Given Parameters

1. Jan 30, 2012

### velouria131

1. The problem statement, all variables and given/known data

A projectile is launched with a speed v at an angle theta above the horizontal. Ignore air resistance. Derive an expression for the angle theta in terms of the parameters of the problem such that the horizontal distance from the launch of the object is N times greater than the maximum height achieved during its flight. Assume that the object lands at the same vertical position from which it was launched and that the acceleration due to gravity is taken as g

2. Relevant equations
v = vo + at
vav = ½(vo + v)
x = xo + ½(vo + v)t
x = xo + vot + ½at²
v² = vo² + 2a(x - xo)
x = xo + vt - ½at²

3. The attempt at a solution
First of all, thanks for any help and hello. I have sat here for quite awhile deliberating, and I seem to just convolute any potential solutions. The first thing I did was draw vector R at angle theta from the horizontal. I then made two tables; table X and table Y. I defined the initial velocity in the X direction as vcos? and the initial velocity in the Y direction as vsin?. I defined the acc. in the Y direction as -9.8 m/s^2. I assumed from this point that I would have to construct an equation - inverse tangent vsin? / vcos?, in terms of vsin? and vcos? that would end of being numerical. I cannot for the life of me figure this part out. The key seems to be that the horizontal distance traveled is N times the maximum vertical distance, which would be N times the the distance traveled by the object whose finally velocity is zero and initial velocity vsin?. Where do I go from here? Or am I even on the correct track?

Thanks again.

Last edited: Jan 30, 2012
2. Jan 30, 2012

### Simon Bridge

When you get lost - draw the v-t diagrams.
For this problem, there are two. One for vertical and one for horizontal motion.

Horizontal is just flat for some time T (leave it as T for now).
The horizontal distance traveled is the area under this graph - horizontal speed times T.
note: if θ is the angle between the initial velocity vector and the ground, then vx=v.cos(θ) ... that's correct :)

The vertical one is a line starting at vy and dropping to -vy in time T. The total area will be zero, which is because you start and finish at the same height. The area under the first half of the graph is the maximum vertical height.

So now your relationship between max height and horizontal distance is actually a relationship between two areas... the area of a triangle and the area of a rectangle - you can do those!

You can get the expression of T from the slope of the vy-t graph (hint: it's -g)

3. Jan 31, 2012

### velouria131

I appreciate the help, that clears up quite a bit. I have just woken up, and upon reading this recognize how expressions may be constructed representative of the angle theta using these two vt graphs. However, will this enable me to solve for an actual angle? Or by expression in terms of the parameter, is the problem simply asking for an equation?

4. Jan 31, 2012

### Simon Bridge

When you write the expressions down in a list, you'll see what needs to be done.
You'll end up with 4 equations and 4 unknowns - hint: sin/cos=tan; leave everything as letters.

5. Jan 31, 2012

### velouria131

Alright, let's see. You say there are four expressions? T = sqrt of (-g)^2 -v^2(sinx)^2. T * vcosx is the horizontal distance and 1/2*T*vsinx is the maximum vertical distance. Am I missing something? Now, I guess the problem I am having (at this point) is conceptualizing as to what I am supposed to do with these expressions (if they are correct that is)! Do I derive expressions for vsinx and vcosx and then take the arctan of this expression? I guess I am lost as to what my ends are here. Thanks.

6. Jan 31, 2012

### Simon Bridge

I don't think that first one is right somehow. You have T=some stuff, then say that is a horizontal distance? The "some stuff" does not have the right units for distance or time.

Put max vertical distance as "h" and the max horizontal distance as "R", and initial speed as "v", initial angle is θ (click "go advanced", or "quote" and look on the right) it will be easier.

From vy-t graph you have an expression for the area of one triangle (there are two) and an expression for the slope of the line (2 equations). From the vx-t graph you have an expression for the area of the whole graph (1 equation). From the description of the problem you have an expression relating max height to range. This one, with the three from the graphs makes four.

The three expressions from the graphs describe every possible motion that has that sort-of shape.
The last expression selects one of the possible motions suggested by the graphs.
So all four of them, together, describe the motion you want.

I'll start you off:
from vx-t graph, R is the area, which is "base times height", so: R=vTcosθ ...eq.(1)
from vy-t graph, h is the area under one triangle, which is "half-base-times-height".
The base is (T/2) and the height is v.sinθ so: 2h=vT.sinθ ...eq.(2)
The slope of the vy-t graph is "rise-over-run" ... you do it. That's eq.(3)
Eq.(4) comes from the relationship between h and R. You do it.

Once you have them, you have completely described the words in the problem as maths.
At that point, you must extract only the parts of the math that is of interest. You need to use algebra to put what you want to know only in terms of things you already know.
So you are not told the max height or the range, but you are told how they relate to each other - so N can be treated as a known quantity. Acceleration of gravity is known. Anything else?

When you have eliminated all your unknowns except for θ, you solve for θ and you have your answer.

Last edited: Jan 31, 2012
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