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!

Please help with free-fall acceleration

  1. Sep 12, 2004 #1
    A rocket on the ground, accelerates straight upward from rest with constant net acceleration "a" , until time "t" , when the fuel is depleted. Here "g" is a positive number equal to the magnitude of the acceleration "a" due to gravity.

    -What is the maximum height reached in terms of a, t and/or g?

    i've been stuck on this for awhile so any help will be greatly appreciated
     
  2. jcsd
  3. Sep 12, 2004 #2

    Pyrrhus

    User Avatar
    Homework Helper

    A lot of students ask this problem... :smile:

    Show me what have you done so far? what is your interpretation?
     
  4. Sep 12, 2004 #3
    i've come to see that the initial velocity is 0, the velocity of the rocket when it runs out of fuel is a*t which i got from the equation v = v_0 + at.

    I'm guessing the maximum height is the distance it travels from the ground to the point where it runs out of fuel plus the distance from that point to its maximum height from free-fall acceleration.

    So from the equation v^2 = V_0^2 + 2a(y) i got the distance from the ground to the point where it runs out of fuel to be a*t^2/2. This is as far as i have gotten and i am not sure how to find the other distance.
     
  5. Sep 12, 2004 #4

    Pyrrhus

    User Avatar
    Homework Helper

    Well, the rocket's final speed when it's fuel is depleted will be the initial speed when it begins Free Fall with g acceleration and its final speed will be 0 when it reaches its max height.
     
  6. Sep 12, 2004 #5
    so would the distance traveled to the max height during free fall acceleration be : 0= ((at^2/2)^2) +gy, and then y = -(a^2t^4)/4g?

    and then would the maximum height be: (at^2)/2 - (a^2t^4)/4g ?

    it doesn't look right, am i doing something wrong?
     
  7. Sep 12, 2004 #6
    does it have anything to do with the fact that the problem stated that g is positive in this problem? :confused:
     
  8. Sep 12, 2004 #7

    Pyrrhus

    User Avatar
    Homework Helper

    Well the problem states g is a positive number equal to the magnitude of the acceleration a, ok. Let's make up negative and down positive then, so g is pointing the positive way and a will be pointing the negative way, also g magnitude will be equal to a.

    Vector magnitudes are always positive, so i'm sure i gave the right interpretation.. if not let me know, my english is not that great.

    now let's divide the problem in two parts, the one with acceleration -a and the other with acceleration g

    Acceleration -a:

    info:
    [tex] V_{o} = 0 [/tex]
    [tex] Y_{o} = 0 [/tex]

    Using the equation:
    [tex] V = V_{o} + at [/tex]
    [tex] V = -at [/tex]

    and:
    [tex] Y - Y_{o} = V_{o}t + \frac{1}{2}at^2 [/tex]
    [tex] Y = - \frac{1}{2}at^2 [/tex]

    Now let's work the second part

    Acceleration g
    info:
    [tex] V_{o} = -at [/tex]
    [tex] Y_{o} = - \frac{1}{2}at^2 [/tex]
    [tex] V = 0 [/tex] at max height

    Using the equation
    [tex] V^2 = V_{o}^2 + 2a(Y-Y_{o}) [/tex]
    [tex] 0 = (-at)^2 + 2g(Y_{max}+\frac{1}{2}at^2) [/tex]

    [tex] Y_{max} = -\frac{a^2t^2}{2g} - \frac{1}{2}at^2 [/tex]
     
    Last edited: Sep 12, 2004
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?