Solving Air Resistance in Projectile Motion

[uchicago2012]

Homework Statement

Suppose we attempt to account for air resistance in our projectile motion in the following (incorrect) way: we alter g so that the acceleration in the y direction is -8 m/s², and introduce a horizontal acceleration of -3 m/s². With these changes, find the landing point of a projectile fired with initial speed 32 m/s at an angle of 25°.

Homework Equations

x - x₀ = (v₀ cos theta₀)t + 1/2 at² (note this is changed from the textbook definition- by the addition of the term (+ 1/2 at²)- bc of the introduction of horizontal acceleration)
y - y₀ = (v₀ sin theta₀)t + 1/2 at²
v_y = v₀ sin theta₀ + at
v_y² = (v₀ sin theta₀)² + 2a(y - y₀)
and other constant acceleration equations altered for the purposes of projectile motion
for constant acceleration:
R = (2v₀²/g) (sin 2theta₀)
where R = the horizontal range of the projectile

The Attempt at a Solution

I was wondering if there was a way to solve this without assuming the landing point is the same as the launch point (constant elevation.) I solved it assuming constant elevation, but as it doesn't say that, I don't know if that was perhaps a bad assumption.

With my assumption, I found that the projectile's landing point was 80.9 m from its starting point.

 

[cepheid]

Homework Equations

x - x₀ = (v₀ cos theta₀)t + 1/2 at² (note this is changed from the textbook definition- by the addition of the term (+ 1/2 at²)- bc of the introduction of horizontal acceleration)
y - y₀ = (v₀ sin theta₀)t + 1/2 at²
v_y = v₀ sin theta₀ + at
v_y² = (v₀ sin theta₀)² + 2a(y - y₀)

These equations are good, but here's a minor nitpick: you should probably explicitly indicate that the acceleration in the equation for x(t) is a_x and the acceleration in the equation for y(t) is a_y. Using the same symbol implies that they are the same quantity.

The Attempt at a Solution

I was wondering if there was a way to solve this without assuming the landing point is the same as the launch point (constant elevation.) I solved it assuming constant elevation, but as it doesn't say that, I don't know if that was perhaps a bad assumption.

One way you could do it, in the general case, is if you were given the profile of the land. So instead of y₀ being a constant, it is function of x and this function defines the elevation profile. You would also have the function y(x) which tells you the trajectory of the object (you can solve for y(x) by eliminating t from the equations for x(t) and y(t)). The landing point would then be intersection of of y(x) and y₀(x). In this case, since you aren't given any information about the elevation profile, it seems sensible to assume the terrain is flat.

 

[uchicago2012]

I thought I'd try it out of interest. This is probably a foolish question:

Is there a way to solve x-x₀ = (v₀ cos theta₀)t + 1/2 a_xt² for t? I can't seem to make the algebra work. I let x₀ = 0, but with the remaining x term, I can't get the equation simplified.

 

[cepheid]

I thought I'd try it out of interest. This is probably a foolish question:

Is there a way to solve x-x₀ = (v₀ cos theta₀)t + 1/2 a_xt² for t? I can't seem to make the algebra work. I let x₀ = 0, but with the remaining x term, I can't get the equation simplified.

The expression is a quadratic in t, therefore you would have to use the quadratic formula in order to solve it.

 

[uchicago2012]

So I arrived at

3/2 m/s² +/- [ (-3/2 m/s² + (116 m/s)x)/(58 m/s)]^1/2 = 3.375 seconds, which makes since, because the left side of the equation is the t term and thus the right side should be in seconds. However, from this point, the equation becomes tedious and confusing. It doesn't make sense to subtract 3/2 m/s² from 3.375 seconds and ignoring it and hoping for an opportunity to simplify later didn't pan out as I had to square the quantity to continue simplification.

 

[cepheid]

Once you've solved for t, if every term on the right hand side doesn't have units of time, then you have done something wrong. My advice would be not to substitute in numbers and units until the end. That just clutters things up (it also makes it hard for me to check your work). Start again with the equation:

\frac{1}{2}a_x t^2 + (v_0 \cos \theta) t - \Delta x = 0​

This equation is of the form:

A t^2 + B t + C = 0​

Therefore, the quadratic formula says:

t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}​

Now you have to express A, B, and C in terms of the original quantities and simplify. The algebra is tedious, to be sure.

 

[uchicago2012]

I reworked it but still ran into the same problem:

x = [(2 (v₀ cos theta) [ (-v₀ sin theta)/(1/2 a_y) + 1/2 a_x ]² - 1/2 a_x ] / [ 4(v₀ cos theta)]

The problem arises when I add the squared term, [ (-v₀ sin theta)/(1/2 a_y) + 1/2 a_x ]². (m/s) / (m/s²) = seconds, then it's seconds plus m/s²