Rocket Motion Problem: Finding Maximum Speed and Height at Burnout

[pentazoid]

Homework Statement

A rocket of initial mas M, of which M-m is fuel, burns its fuel at a constant rate in time tau and ejects the exhausts gases with constant speed u. The rocket starts from rest and moves vertically under uniform gravity . Show that the maximum speed achieved by the rocket is u ln(\gamma)-g\tau and that its height at burnout is

u\tau(1-ln(\gamma)/(\gamma-1) where \gamma=M/m[assume that the thrust is such that the rocket takes off immediately.)

Homework Equations

The Attempt at a Solution

I had no trouble finding v, I had trouble integrating v to obtain the height. v=u ln (gamma)-g*tau . h=\intv dt= \intu*ln(m₀/m(t))-.5*gt^2

u is treated as a constant I think since I am integrating v with respect to dt. \intln(\gamma)=\gamma*ln(\gamma)-\gamma. Now I am stuck on this part of the solution.

 

[chrisk]

You are integrating a ln function. The integral is

int[ln(ax)] = xln(ax) - x.

The exhaust velocity, u, is treated as a constant. This integral is obtained by integrating by parts. An introductory calculus text will have the derivation for the intergral of ln(ax). Apply the limits of integration to each part of the solution.

 

[chrisk]

Sorry about the previous post. Your equation for v is correct. Since v = dx/dt, separate the variables such that dx is on the left side and dt is on the right side. Recall, you are finding the maximum height so integrate dx from x = 0 to x = h and integrate (u ln (gamma) - gt)dt from t = 0 to t = tau. Then the remainder of the rocket flight is only under the force of gravity and becomes a vertically fired projectile with an initial velocity. Use v² = vo² - 2g(x - xo) to find the additional height by setting v² = 0. Add this height, x - xo to the height at burnout.

 

[pentazoid]

Sorry about the previous post. Your equation for v is correct. Since v = dx/dt, separate the variables such that dx is on the left side and dt is on the right side. Recall, you are finding the maximum height so integrate dx from x = 0 to x = h and integrate (u ln (gamma) - gt)dt from t = 0 to t = tau. Then the remainder of the rocket flight is only under the force of gravity and becomes a vertically fired projectile with an initial velocity. Use v² = vo² - 2g(x - xo) to find the additional height by setting v² = 0. Add this height, x - xo to the height at burnout.

so I should still integrate (u ln (gamma)-gt)dt I am okay with that part of the problem and I am okay with integrating ln (gamma) because I can integrate ln(gamma) using integration by parts. How would the equation v² = vo² - 2g(x - xo) assist me in helping me find the height? Why do I even need the equation for v^2 for this problem?

 

[chrisk]

The additional equation v² = ... is used because the rocket doesn't suddenly stop once the fuel is gone.

 

[pentazoid]

The additional equation v² = ... is used because the rocket doesn't suddenly stop once the fuel is gone.

I am integrating u*ln (gamma)-gt)dt to obtain the height. I don't see any other use for the additional equation. couldn't I just plug in my initial conditions in u*ln (gamma)-gt)dt to find the maximum height?

 

[chrisk]

Sorry, you do not need the additional equation. I missed the part "max height at burnout". Yes, just evaluate the integral using the initial conditions for height and time and the final time tau.