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Homework Help: Projectile motion with air resistance

  1. May 24, 2015 #1
    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


    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. jcsd
  3. May 24, 2015 #2
    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
  4. May 24, 2015 #3
    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!
  5. May 24, 2015 #4
    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! :)
    $$x =\tau v_{x0}(1-e^{\frac{-t}{\tau}})$$
  6. May 24, 2015 #5
    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))##
  7. May 25, 2015 #6
    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!
  8. May 25, 2015 #7
    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}}.##
  9. May 25, 2015 #8
    After I post it, I find it containing lots of errors... Maybe just regard it as reference and let me check again!!
  10. May 25, 2015 #9
    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}.##
  11. May 25, 2015 #10
    #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 $$y\approx\tau(v_{y0}-g\tau)(\frac{t}{\tau})+gt\tau=v_{y0}t.$$
  12. May 25, 2015 #11


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    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.
  13. May 25, 2015 #12
    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
  14. May 25, 2015 #13
    Yes, I got y for the same way as yours.
    Is the relation I wrote what the problem require?
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