1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Rocket Height (Motion in streight line)

  1. Jan 19, 2008 #1
    [SOLVED] Rocket Height (Motion in streight line)

    Hey i'm 75% sure i have the correct answer, but i have to turn it in tomorrow and it's a lot of points so i just want to make sure i'm going it right.

    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 49.0 . The acceleration period lasts for time 9.00 until the fuel is exhausted. After that, the rocket is in free fall.

    Find the maximum height ymax reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.8 m/s^2 .

    2. Relevant equations
    Sorry if this doesn't look so hot, the formulas were given to me with lost of sub things and i don't know how to make them look good on forums, i'm trying the Latex for the first time

    y(t)= y[tex]_{i0}[/tex] + v[tex]_{i0}[/tex]t + (at[tex]^{2}[/tex])/2
    v(t)=v[tex]_{i0}[/tex] + at

    The i0 should be at the bottom and it means initial at time zero but you probably already knew that.

    3. The attempt at a solution

    y(t)= y[tex]_{i0}[/tex] + v[tex]_{i0}[/tex]t + (at[tex]^{2}[/tex])/2 for 0 < t < 9
    Declare variables and set them equal to values in the problem.
    y[tex]_{i0}[/tex] = 0
    v[tex]_{i0}[/tex] = 0
    a = 49
    formula for postion after 9 seconds of acceleration.

    solve to a number = 1984.5 m at t = 9

    Now calculate the velocity at time = 9.
    v(t)=v[tex]_{i0}[/tex] + at
    v(9)=0 + 49(9)
    v(9) = 441 m/s

    Create a new position fuctionfor time after 9 to the time where the rocket is at 1984.5.

    y(t)= y[tex]_{i0}[/tex] + v[tex]_{i0}[/tex]t + (at[tex]^{2}[/tex])/2 for 9 < t
    Declate Variables
    y[tex]_{i0}[/tex] = 1984.5m
    v[tex]_{i0}[/tex] = 441 m/s
    a = -9.8
    t = (t-9) (offset time to account for the first part)

    y(t)= 1984.5 + 441(t-9) + (-9.8(t-9)[tex]^{2}[/tex])/2

    I don't know how to do this properly, but i know if i sub in an x for (t-9) and solve for the x intercepts and then add them and divide by two i will get the time that it's at the top, and then plug that time back into the function and i should get the answer. to avoid the quadratic forumla i'm putting it into grapher on my mac.

    and get something around 3*10^5 ?
    Last edited: Jan 19, 2008
  2. jcsd
  3. Jan 19, 2008 #2


    User Avatar
    Homework Helper

    What is the velocity of the rocket at maximum height? Using this information you can work out the time the rocket takes to go from when the fuel runs out to when it reaches maximum height. Then you can find the distance it goes in this time.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook