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Homework Help: Kinematics - Maximum altitude

  1. Oct 5, 2013 #1
    1. The problem statement, all variables and given/known data

    A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the rocket. The bolt hits the ground 8 seconds later. Three seconds after the bolt falls off, the rocket's engines fail. How high does the rocket go before it begins to fall back to earth?

    2. Relevant equations
    [tex]x = x_0 + v_0 t + (1/2) a t^2[/tex]

    [tex]v = v_0 + a t[/tex]

    3. The attempt at a solution
    I used the first equation to find the distance between launch and the bolt falling off. I got 313.6m

    After that I used the values I had to find the acceleration until the engine failed. 39.2m/s^2 ?

    Then I just calculated distances, velocities, and times with the values I had. Ultimately I found a maximum height of 3601.5m. For whatever reason this just doesn't seem right. At the moment the engine failed I found a height of 646m. For the rocket to travel that far with a negative acceleration doesn't seem kosher. Any help would be great. Thanks.
  2. jcsd
  3. Oct 5, 2013 #2


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    Staff: Mentor

    I agree with you, your numbers look off.

    Can you show your work in detail?
  4. Oct 5, 2013 #3
    Call [tex]h_1,v_1,t_1[/tex] the height, velocity and the time at the point where the bolt comes off.


    $$v_0=0, h_0=0$$


    The height of the bolt after it comes off will be

    $$h=h_1+v_1 *t-0.5gt^2$$

    You seem to have forgotten the v_1 term or something.
  5. Oct 5, 2013 #4
    I set up an event diagram with 4 events. Y_0 is launch, Y_1 is bolt falling off, Y_2 is engine failure, and Y_3 is max altitude.
    To calculate the distance between Y_0 and Y_1 I used the time it took for the bolt to hit the ground (8s).
    Y_1= .5(-9.8)(8)^2=313.6m
    Then to find the acceleration I used 313.6m=.5(a)(4)^2 and got 39.2m/s^2. Since the acceleration is constant until the engine fails this is the number I'm using for acceleration up until that point.
    To find velocity(v_1) at Y_1 I used v_1=39.2(4)=156.8m/s
    Then Y_2=156.8(3)+.5(39.2)(3)^2 = 646.8 +Y_1= 960.2m (this is a correction)
    Velocity(v_2) at Y_2 I used v_2=156.8(3)= 274.4m/s
    To find time at max altitude t=(-274.4)/(-9.8)=28seconds
    Finally Y_3=274.4(21)+.5(-9.8)(21)^2 +Y_2= 4561.7 m ( this is also a correction)
  6. Oct 5, 2013 #5
    No. This is the motion of a particle that is dropped from REST at 313.6m height.
  7. Oct 5, 2013 #6
    When the bolt detaches from the rocket, it will still have a velocity upwards (in the beginning). It does not suddenly come to a complete stop and then start falling for 8 seconds
  8. Oct 5, 2013 #7


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    Staff: Mentor

    When the bolt fell off it didn't start falling from rest. It had whatever upward speed the rocket had at the moment it detached. What you've calculated is the height for an object that starts from rest and takes 8 seconds to reach the ground.

    You'll need to include expressions for the rocket speed and height at the moment the bolt detaches based on the yet-to-be-found value of a. You can then solve for a by the method you've used, there will just be a couple more terms with 'a' in them in your bolt trajectory equation.
  9. Oct 5, 2013 #8
    I think I'm following you. For example, would my v_1= 4a? my Y_1=8a? Am I on the right track?
  10. Oct 5, 2013 #9


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    Staff: Mentor

    Yes indeedy :smile: Continue!
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