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Crazy Projectile Motion Problem Coming Right At Yah!

  1. Oct 7, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the initial velocity and the maximum height of the rocket.

    Angle = 90˚
    Horizontal distance = 0 m
    Time = 6.55 seconds

    2. Relevant equations

    Suvat equations.

    3. The attempt at a solution

    s = ut + 1/2a(t^2)
    s = 6.55u
    0/6.55 = u

    ???

    Please help!
     
  2. jcsd
  3. Oct 7, 2011 #2
    It would help to write down the known quantities and the unknowns:

    v, t and a are known. Since the angle is 90°, there is no horizontal component involved. The unknowns are u and S. Thinking of equations where only one of the unknowns is used will help.
     
  4. Oct 7, 2011 #3
    v = u + at?

    but how does that work I would get

    0 = 0 + 0(6.55)
     
  5. Oct 7, 2011 #4
    If both initial and final velocities are zero and the acceleration is zero, how does the rocket move?

    u cannot be zero since that is what you are supposed to find . What is the velocity (final velocity) of the rocket at its maximum height? What is its acceleration?
     
  6. Oct 7, 2011 #5
    Thanks so far i got u which is 32.75 I think but I don't know how to find the maximum height.
     
  7. Oct 7, 2011 #6
    Funny that they call it a rocket. Rockets are usually self propulsive by hurling mass backwards. This is a normal projectile, right? :)
     
  8. Oct 7, 2011 #7
    s = vt + 1/2a(t^2).

    here you have a solution for s. if you plug in s=0 (the ground where the projectile starts), you probably have two different t that satisfy this. One is obviously t=0, but the other?.

    As for the maximum height; well, what you have here is distance as a function of time s(t). What about finding the maximum value of this graph? ;)
     
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