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Maximum Initial Velocity

  1. Feb 6, 2016 #1
    1. The problem statement, all variables and given/known data
    A German U2 rocket from WW2 had a range of 300km, reaching a maximum height of 100km. Determine the rocket's maximum intial velocity.

    X component variables: di=0, df=300, a=0, time=9, vi=?
    Y component variables: a=-9.8, vf=0, vi=?, di=0, df=100, displacement=100, Note: For the y components, I counted the maxium height as the final position and the velocity at the max height as the final velocity.

    2. Relevant equations
    The big 5 kinematics equations
    3. The attempt at a solution
    I attempted to find the initial velocity for the x and y components and then use pythagorean theorem to solve for the magnitude of the max velocity as you would solve for the hypotenuse of a triangle and then I would use the cosine law to solve for the angle at which the rocket was launched. But, I got the incorrect magnitude after doing pythagorean theorem, so I stopped. Here's what I did:

    First, I calculated initial velocity for the y component, using the equation vf^2=vi^2+2a(displacement) --- I got vi=~44.3m/s. Then, I used another equation with the y component variables to solve for time because I needed another variable to use for the x component equations. For finding the time, I used displacement=1/2(vf+vi)(time). I got time=4.5s, but since the variables I used were variables only from the initial position to the max height, I think that using those variables only solved for the time up to the max height, so I multiplied by 2 to get the time for the whole motion--- 9 s.


    Then, I used the equation df=di+vi(time)+1/2a(t)^2. I got vi=33.3.


    Then, I used pythagorean theorem with the magnitudes of my x and y component velocities---- a^2=b^2-c^2 -- I plugged in the x and y component velocities for b and c and solved for a. I got a=55.4. This should've been equal to the magnitude of the max velocity, but the book says the answer is 1.75 x 10^3 m/s. Can someone tell me where I went wrong.
     
  2. jcsd
  3. Feb 6, 2016 #2

    SteamKing

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    For one thing, in your initial calculations, you apparently used a range of 300 meters, instead of 300 kilometers.
    Ditto for the maximum altitude; you used 100 meters instead of 100 kilometers.

    It makes a difference, those pesky units. :rolleyes:


    BTW, the name of the rocket was the V-2, not the U-2. The U-2 was a famous U.S. jet spy plane.
     
  4. Feb 6, 2016 #3
    My textbook says U2, but okay. And thank you.
     
  5. Feb 6, 2016 #4

    SteamKing

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  6. Feb 6, 2016 #5
    Was I right about multiplying the time by 2 for the y component?
     
  7. Feb 6, 2016 #6

    SteamKing

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    You got bigger fish to fry than that. Did you not read the part about using the wrong units in your calculations?
     
  8. Feb 6, 2016 #7
    Yes, but I can easily fix that, the time would also affect my calculations
     
  9. Feb 6, 2016 #8

    SteamKing

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    Once the rocket reaches maximum altitude, it's no longer powered. You can calculate the time it takes to free-fall back to earth.
     
  10. Feb 6, 2016 #9
    Don't you just multiply the time it took to reach the max height by 2 to get the time for the entire motion. That's what I did, and then I used that time value in my calculations for the x component.
     
  11. Feb 6, 2016 #10

    SteamKing

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    At launch, the rocket has zero initial velocity, and it takes time for the rocket to build up to its maximum velocity before the engine cuts off. That's why I suggested calculating the time to free fall from the highest altitude, since it's not clear the velocity profile going up while powered will be the same one coming down in free-fall.
     
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