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2-D Kinematic Question

  1. Jul 12, 2007 #1
    A group of people are watching a fireworks display. 3.5 s after they hear a rocket launch, they see the explosion. the explosion is observed to be located 50.0 from the launch point at an angle of 73.0 degrees above the horizontal.

    (a) Sketch the problem. Label all parts of the problem with the appropriate variables. This includes all variables you will use in your equations. Specify the value of all the variables you are given.

    (b) What is the initial velocity of the rocket?

    (c) What is the velocity of the rocket just before it explodes?

    2. Relevant equations

    v = v0 + at

    3. The attempt at a solution

    First I have a couple of questions.

    Is the rocket accelerating? If yes, is it in the x and y directions?

    Is there an assumption I should be making to solve this problem?

    cos73 = x/50m
    x= cos73(50m)

    sin 73 = y/50m

    v = v0 + at
    0 = v0 + (-9.80)(3.5s)
    vy0 = 34.3 m/s

    v=4.18 m/s

    Then I would use the pythagorean theorem to solve for the final velocity but this is all done assuming that the velocity was constant and that the rocket was not accelerating.
  2. jcsd
  3. Jul 12, 2007 #2
    bump bump ////
  4. Jul 12, 2007 #3


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    the rocket is accelerating under the influence of gravity, so d=vt is not an equation you want to use. use the equations which are relevant for the case of uniform acceleration in the negative y-direction (at 9.8m/s^2) and no acceleration in the x-direction.
  5. Jul 13, 2007 #4


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    If we assume that the fuel is spent to launch the rocket then we are dealing with parabolic motion.

    Note that the initial speed, [tex]V[/tex], can be calculated from the launching angle, [tex]\theta _o[/tex], and the horizontal speed, [tex]V_x[/tex] (which can be calculated from the given data - I would think that the given distance is actually the horizontal distance covered up to the point of explosion)

    [tex]V = \frac{V_x}{\cos(\theta _o)}[/tex]

    to determine the initial angle use the known distances covered in the given time

    [tex]x = Vt\cos(\theta _o)[/tex] ..... (1)


    [tex]y = Vt\sin(\theta _o) - \frac{1}{2}gt^2[/tex]

    transferring the last term (which can be calculated) to the left gives

    [tex]y + \frac{1}{2}gt^2 = Vt\sin(\theta _o)[/tex] ..... (2)

    dividing (2) by (1) now gives the launching angle [tex]\theta _o[/tex].
    Last edited: Jul 13, 2007

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