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: Height of a rocket at a certain time

  1. Jan 19, 2009 #1
    1. The problem statement, all variables and given/known data

    During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.20 m/s^2. When it is 260 m above the launch pad, it discards a used fuel canister by simply disconnecting it. Once it is disconnected, the only force acting on the canister is gravity (air resistance can be ignored).

    How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration?

    2. Relevant equations

    1. v^2 = v(0)^2 - 2g(y - y(0))
    2. y = y(0) + v(0)t - (1/2)at^2
    3. v = v(0) - gt

    3. The attempt at a solution

    first i wanted to find the speed of the rocket at 260m by using equation 1. and solving for v. I attained 40.8m/s.

    then I used this 40.8m/s as the initial velocity of the canister going up and v = 0 (speed of the canister at its maximum height) as the final velocity to find the initial height of the canister before it begins to drop. I attained 345m.

    once i found this i wanted to find the final velocity of the canister before it hit the pad and used equation 1 again. I attained v = 82.2m/s.

    after, i found the time it took the canister to hit the ground by using equation 3. I attained 8.39s.

    finally, i used this time interval along with v(0) = 40.8m/s, y(0) = 260m, a = 3.2m/s^2 in equation 2. I attained 715m which is wrong according to the course website.

    can anybody help me?
    Last edited: Jan 19, 2009
  2. jcsd
  3. Jan 19, 2009 #2


    User Avatar
    Homework Helper

    By your method there are 3 time intervals that need to be determined.

    1. time to 260 m
    2. time to canister max
    3. time to free fall.

    40.8 m/s / 3.2 yields your first time.
    40.8 m/s / 9.8 yields your second time
    345m = 1/2*9.8*t2 yields your third.

    Add all 3 together and that put back into 1/2*(3.2)*t2 should yield your rocket height right?

    Alternatively you can plug the 40.8 and the 260 into the initial conditions relating time, distance,and velocity and solve the quadratic
  4. Jan 20, 2009 #3
    wow! Thank you so much LowlyPion. I did not notice how there were more than one time intervals.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook