Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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? ;)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook