# Homework Help: Projectile motion with air resistance

1. May 24, 2015

### ma18

1. The problem statement, all variables and given/known data
This is a standard projectile motion problem, the mass is m, the drag is F_r = - alpha *v, where alpha is a positive retarding coefficient. The origin is the ground and at time t = 0 the horizontal and vertical velocities are positive

a) Write down Newton's second law for the horizontal component of the velocity, v_x (t). Solve for v_x (t) with given initial conditions and use the characteristic time, tau = m/alpha in your answer. Give physical reasoning as to why m increases with tau and why alpha decreases with tau.

b) Integrate v_x (t) from a to determine x(t) with given initial conditions. What is x(t) with the limit t goes to infinity? What would be the horizontal position at t = tau?

c) Convert the DE from a into an equation for v_x (x). Solve for v_x (x) with the given initial conditions and use tau in your answer.

2. Relevant equations

F = ma

3. The attempt at a solution

a)

F_x = m*v'_x = - alpha*v_x

v'_x/v_x = d/dt *ln(v_x) = - alpha * v_x

ln v_x = - alpha/m t + ln v_x0

v_x (t) = v_x0 * e^(-t/tau)

b) I am stuck here, I integrate to get

integral v_x (t) = x(t) = v_x0 *e^(1-t/tau) / (1-t/tau) * A (where A is an arbitrary constant of integration)

But solving for A with the condition that x(t=0) = 0 I get A = 0

c) I am stuck here too,

m dv_x/dx * dx/dt = - alpha * v(t)

I don't know how the convert the second v(t) into a v(x)

Any help would be much appreciated.

2. May 24, 2015

### tommyxu3

Hello, I'm a little confused. The relation $$dx/dt = v_x = v_{x0} e^{(-t/\tau)}$$ is confirmed. After integral, I got $$x = -\tau v_{x0} e^{(-t/\tau)} + c$$ while c is a constant. Then I can get $c=\tau v_{x0}$. Does it work?

Last edited: May 24, 2015
3. May 24, 2015

### ma18

Oh, I see I did the integral wrong, thanks.

But when I put it in for x(t=0) = 0 I get:

0 = - tau *v_x0 * e^(0) + c
c = tau *v_x0

Not c=0, but that is fine.

Can you help me with c)?

Thanks a lot!

4. May 24, 2015

### tommyxu3

I typed $c=0$ at first and found it wrong just later and I thought you hadn't seen it yet...
OK, I have corrected it and it should be what you said! :)
$$x =\tau v_{x0}(1-e^{\frac{-t}{\tau}})$$

5. May 24, 2015

### tommyxu3

For the convert, based on what we have known: $$x =\tau v_{x0}(1-e^{\frac{-t}{\tau}}),v_x=v_{x0}e^{\frac{-t}{\tau}} ,$$we can get $e^{\frac{-t}{\tau}}$ in $x$ and let it into $v_{x}$ so we get $$v_x=v_{x0} (1-\frac{x}{\tau v_{x0}}).$$
Does this what the problem require? $(v_{x0}(x))$

6. May 25, 2015

### ma18

I think it expects you arrive at the answer a different way but that is still correct.

There is also other parts to this question

d) Write down the DE for the vertical portion and solve for v_y with the given initial conditions

I got this to be v_y = (v_y0 - g*tau) e^(-t/tau) + g*tau and this is correct.

e) Determine t_a, the time at which the projectile reaches max height. As alpha --> 0, tau --> infinity. Use ln(1+x) = x - x^2/2+... for x<1 to determine lim (tau --> infinity) t_a

For this I set v_y = 0 as it would be at the max time, then rearranged and solved for t:

(v_y0 - g*tau) e^(-t/tau) + g*tau = 0

e^(-t/tau) = -g*tau/(v_y0 - g*tau)

t = t_a = -tau ln (-g*tau/(v_y0 - g*tau))

Taking the limit as tau ---> infinity

Using L.H's rule the term inside the ln goes to 1, and thus ln(1) goes to 0 but the tau goes to infinity, so t_a goes to 0. However, I know I messed this up somewhere, can you see my mistake?

f) Determine y(t) by integrating v_y(t). Use the fact that e^x = 1 + x + x^2/2+... to determine lim t-< infinity y(t)

So for the integration I got:

y(t) = tau e^(-t/tau) (g*tau-v_y0)+g*t*tau

Using the property of e^x

tau * e ^ (t/tau) = tau (1 - t/tau + t^2/tau^2) = tau - t + t^2/tau

The last term goes to zero as tau goes to infinity.

So we get

y(t) = (tau - t) (g*tau-v)+gt*tau
= g*tau^2

which goes to infinity.

This is wrong as well but again I am not sure where my mistake is.

Any help would be much appreciated again, thanks a lot!

7. May 25, 2015

### tommyxu3

For (d) I got a little difference from yours... mine is $$v_y=(v_{y0}+g\tau)e^{(\frac{-t}{\tau})}-g\tau,$$ but this time I cannot find my wrong turn while yours is correct...
Let me regard yours is correct, and then we get $$v_y=(v_{y0}-g\tau)e^{(\frac{-t}{\tau})}+g\tau.$$Solve t and get $$t_a=-\tau \ln{\frac{g\tau}{-v_{y0}+g\tau}}.$$When $\tau\rightarrow\infty$, $$t_a=-\ln{(1-\frac{g}{v_{y0}}\tau)^\tau}$$ $$\rightarrow-\ln{e^{\frac{-g}{v_{y0}}}}=\frac{g}{v_{y0}}.$$
So, it may be $\frac{g}{v_{y0}}.$

8. May 25, 2015

### tommyxu3

After I post it, I find it containing lots of errors... Maybe just regard it as reference and let me check again!!

9. May 25, 2015

### tommyxu3

I got it!
When $\tau\rightarrow\infty$, $$t_a=\ln{(1-\frac{v_{y0}}{g\tau})^\tau}$$ $$\rightarrow\ln{e^{\frac{v_{y0}}{g}}}=\frac{v_{y0}}{g}.$$
So, it may be $\frac{v_{y0}}{g}.$

10. May 25, 2015

### tommyxu3

#7 is apparently absurd. Please ignore it! :p
For your $y(t)$, I'm not sure if you neglected the constant of integral. I got $y=\tau(v_{y0}-g\tau)(1-e^{\frac{-t}{\tau}})$ for $y_0=0.$ Then I can get $$y\approx\tau(v_{y0}-g\tau)(\frac{t}{\tau})+gt\tau=v_{y0}t.$$

11. May 25, 2015

### vela

Staff Emeritus
There's nothing to do. The only difference between v(t) and v(x) is which variable you write the velocity in terms of. Either way, it's the velocity. It's probably just a typo, but it should be $v_x(t)$, not $v(t)$, on the righthand side.

12. May 25, 2015

### ma18

How did you arrive at that value for y?

So we have

v_y = (v_y0 - g *tau) e^(-t/tau)+g*tau

Integrating we get

y = - tau * v_y0 * e^(t/tau) + g*tau^2*e^(-t/tau)+gt*tau + C
= tau * e^(-t/tau) (g*tau-v_y0) + gt*tau + C

taking y(t=0) = 0

0 = tau (g*tau-v_y0) + C

=>

C = tau (v_y0 - g*tau)

==>

y = tau*e^(-t/tau) ( g tau - v_y0) + gt tau + tau*v_y0 - g*tau^2

13. May 25, 2015

### tommyxu3

Yes, I got y for the same way as yours.
Is the relation I wrote what the problem require?