1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rocket Height from Launch

  1. Jan 25, 2009 #1
    1. The problem statement, all variables and given/known data

    A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 53.9 . The acceleration period lasts for time 5.00 until the fuel is exhausted. After that, the rocket is in free fall.

    Find the maximum height reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.8 .

    2. Relevant equations

    Distance = 1/2 at^2
    Velocity = u + at

    3. The attempt at a solution

    So far what I get
    distance = 1/2 x 53.9 x 5^2 = 673.75 m
    V = 0 + 53.9 x 5 = 269.5 m/s

    and from there I seem to be stuck. I dont know what else to do, if there is someone to help guide me or of this ditch I'd appreciate it.
  2. jcsd
  3. Jan 25, 2009 #2


    User Avatar
    Homework Helper

    That's a good start.

    Now how much longer until it gets to max height?

    Use the same equation that gives you your speed at main engine shut-off. Only this time it's gravity and not rocket power.

    Armed with that time figure then using your distance equation how much further it went. Add the 2 heights together. Done.
  4. Jan 25, 2009 #3
    so I multiplied 53.9 by 9.8 and got 528.22, then added that with 676.75 to get 1201.97. Thats not the answer, im still off somewhere or mis read?
  5. Jan 25, 2009 #4
    Once its done accelerating.

    Solve for t

    Distance formula
    d0 = 673.75m
    v0= 269.5m/s
  6. Jan 25, 2009 #5


    User Avatar
    Homework Helper

    No. You don't multiply the accelerations together. You use the acceleration of gravity this time instead of the acceleration of the rocket over the 5 seconds.

    You figured velocity is 269.5. Divide that by 9.8. That's how many seconds it will continue to rise until velocity is back to 0 - time to max height.

    Use that time with y = 1/2*g*t2
    to find how much farther it goes against gravity.

    That height is what you add to your previous height at main engine cut off to get total height.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?