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Free falling body after certain height

  1. Aug 22, 2011 #1
    1. The problem statement, all variables and given/known data

    A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25 m/s^2 and feels no appreciable air resistance. When it has reached a height of 525 m, its engines suddenly fail so that the only force acting on it is now gravity.

    (a) What is the maximum height this rocket will reach above the launch pad?

    2. Relevant equations

    [itex]velocity_{y-axis} = v_{0y} + \int \! a_y \, \, dt[/itex]
    [itex]position (y) = y_0 + \int \! v_y \, \, dt[/itex]

    3. The attempt at a solution

    First I list all the given information,

    [itex]a_y = 2.25 m/s^2[/itex]
    [itex]-g = -9.80 m/s^2[/itex]

    Next I integrate [itex]a_y[/itex] in order to obtain the velocity,

    [itex]v_y = \int \! 2.25 m/s^2 \, \, dt[/itex]
    [itex]v_y = 2.25t[/itex]

    Now I integrate [itex]v_y[/itex] in order to obtain the position,

    [itex]y = \int \! 2.25t \, \, dt[/itex]
    [itex]y = 1.125t^2[/itex]

    And this is where I get confused. I solve the following quadratic equation in order to determine the maximum height that was reached (gravity gets differentiated, no?):

    [itex]0 = 1.125t^2 -4.90t + 525[/itex]

    And then I get 525.258 m instead of the correct answer of 646 m. What am I doing wrong?
     
  2. jcsd
  3. Aug 22, 2011 #2

    PeterO

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    Why are you integrating???
    Just use equations of motion to find the speed where the rocket engines fail.
    You then have an initial height, Initial velocity, final velocity and acceleration so you can find the maximum height reached.

    v^2 = u^2 + 2as for velocity at height 525m

    then again v^2 = u^2 + 2as to find how much further up the rocket goes. Don't forget to add this extra height to 525 to get the maximum height.
     
  4. Aug 23, 2011 #3
    **Considering the rocket when it reaches the height of 525m.
    Because upward acceleration is constant:

    Vyf = Vyi + Ayt

    **However to find Vyf you need to know the time it takes to reach 525m.
    Yf = Yi + Vyi + 1/2Ayt2
    [2(Yf)/Ay]1/2=t

    Vyf = Ayt
    Vyf= 48m/s

    **now to find the addition height when the rockets fail
    for linear motion

    ads=vdv

    a[itex]\int[/itex]dy=v[itex]\int[/itex]dv
    **the upper and lower limits of integration for dy : Yf and 0
    ---------for convince we are saying setting the initial position to zero when the rocket has reached 525m in the sky
    **the upper and lower limits of integration for dv : Vyi = 48m/s and Vyf = 0 ( because eventually gravity overtakes the rocket and discontinues upward displacement)

    -9.81(Yf-0) = 0-(482/2)

    Yf = -482/(-9.81*2) = 117m

    hmax = 525m+117m = 642m

    OR YOU CAN DO THIS:
    The velocity of the rocket 48m/s when the engines fail.
    At this point the rocket is experiencing downward acceleration due to gravity.
    So you must find how much further will the rocket vertically displace before the rocket is completely over come by gravity and enters free fall.
    So find the time it takes for the upward velocity to become zero when the rocket engines fail:

    **find the time when the upward velocity is zero
    **because the downward acceleration from gravity is constant

    Vyf = Vyi -gyt
    -Vyi/-g = t = 4.89s

    Yf = Yi + Vyit - 1/2gt2
    **vyi=48m/s
    Yf=117m

    hmax = 525m+117m = 642m
     
  5. Aug 24, 2011 #4

    PeterO

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    If you don't round off the figures on the way through, you will get 646 [ 645.53... actually]

    There is a formula connecting Initial velocity, final velocity, acceleration and displacement

    Vf^2 = Vi^2 + 2*acc*disp. [I know it as v^2 = u^2 + 2as]

    That can be used to calculate how fast the rocket is going when it reaches a height of 525m while accelerating at 2.25

    It can then be used to calculate how much further it travels starting at that speed, and slowing to zero with an acceleration of 9.8 down.

    Add that distance to 525 and you have maximum height. BUT DON'T ROUND OFF ON THE WAY THROUGH THE CALCULATIONS. For the speed at 525m, don't even bother to take the square root - you only have to square it again when you sub into the next formula.
     
  6. Aug 24, 2011 #5
    Just to point out an assumption you're making:

    g isn't even a constant at the surface of the Earth. You're saying that at that height, g will still be the same as higher up, which isn't absolutely true. However, it's not too bad an approximation, and it will save you from some pretty confusing calculus.

    Apart from that, you really only need one equation, the one stated by most people above me. v^2=u^2+2as (where v is final velocity, u is initial, a is acceleration and s is displacement)

    For the first half, we're trying to find out how fast the rocket is going when it stops accelerating due to it's thrusters. v^2= 0+2(2.25*525)

    Then with that answer, we know the maximum height will be when the rocket stops going upwards and starts to come back to Earth. So, we make v=0, a=g and u=the answer we get in the previous calculation, or the one previously called v^2:
    0=(u^2) -2gs
    s=(u^2)/2g

    s will be the extra amount it goes, so add on 525 m and voila! Answer!

    (I get 525m+128 (with lots of decimals), so about 653 metres)
     
  7. Aug 24, 2011 #6
    Thank you all for the wonderful explanations, that really helped me :)
     
  8. Aug 24, 2011 #7

    lightgrav

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    What you did wrong originally, was:
    (1) tried to set the total displacement to zero, instead of the velocity to zero
    (2) used the same variable, "t" , for two distinct intervals (powered and unpowered)
    (3) goofed up with units, treating "½ g t" as a displacement ("t" with no square)

    I'll wager that you did not diagram this scenario, and label the important features.
     
  9. Aug 25, 2011 #8
    Surely one doesn't have to draw a diagram EVERY time! That would be an enormous waste of time!
     
  10. Aug 25, 2011 #9

    PeterO

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    Some people are able to close their eyes and draw the diagram in their mind in about 1 or 2 seconds.
    Most people need to draw a real diagram. 10 seconds of drawing to avoid 60 seconds of wondering what the heck to do, then waiting a few days for your post on Physics Forums to be answered completely.
     
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