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Initial Velocity of a Rocket

  1. Jun 9, 2009 #1
    1. The problem statement, all variables and given/known data
    I need to find the initial velocity of a rocket using the method a=[tex]\Delta[/tex]v/[tex]\Delta[/tex]t. Given a situation where [tex]\Delta[/tex]t = 4 seconds. The force of gravity in this case is -10(m/s)

    2. Relevant equations
    After I get [tex]\Delta[/tex]v how do I turn that into the initial velocity?

    3. The attempt at a solution
    a=[tex]\Delta[/tex] v/[tex]\Delta[/tex] t
    -10=[tex]\Delta[/tex] v / 4 seconds
    [tex]\Delta[/tex] v = -40 (m/s)
  2. jcsd
  3. Jun 9, 2009 #2


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    Homework Helper

    Welcome to PF.

    What you found was the speed of an object dropped after 4 seconds ... or if you are talking about a rocket, the speed in accelerating from rest at 10 m/s2.

    The acceleration of gravity while of interest doesn't say anything about the force the propellant delivers to the motion of a rocket.
  4. Jun 9, 2009 #3
    This is just after launch so I see no need to worry about any force acting on the rocket while it's in the air other than gravity pulling it down.
  5. Jun 9, 2009 #4


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    can you type the full question please, it sounds fuzzy like this
  6. Jun 9, 2009 #5
    A rocket has been in the air for 4 seconds from the instant it was launched to when it hit the ground. The force of gravity can be rounded to 10(m/s). Using the method a= delta v / delta t, find the initial velocity for the rocket.
  7. Jun 9, 2009 #6


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    you mean 10 (m/s^2)

    wouldn't that motion of the rocket be a bow like motion [projectile motion]since it didn't mention that it was verticly launched!

    if so work on the vertical axis, where the final velocity=0 and the intial one v [as you did], then try to find the distance from start point to hit point to get v horizontal which is constant [not accelerated] then add both to get the initial velocity [v_i=squareroot(v_v^2+v_h^2)]
  8. Jun 9, 2009 #7


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    OK. So it is not about the rocket at all except that it is a projectile after launch?

    In which case the usual rules apply.

    V = Vo - a*t = Vo - g*t = Vo - 10*t on the way up and then again on the way down.

    With that in mind, then by symmetry it takes 2 seconds up, and 2 more down.

    At 2 seconds to max height, then initial velocity is ...
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