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Need Help with Projectile Motion Problem

  1. Sep 27, 2007 #1
    1. The problem statement, all variables and given/known data

    I have a problem here and I just can't seem to figure it out. The problem is, A ball is thrown horizontally from a height of 24.28 m and hits the ground with a speed that is 4.0 times its initial speed. What was the initial speed?

    2. The attempt at a solution

    I've found the velocity in the y-direction by using the equation

    v^2 = vo^2 + 2 a x, the velocity is 21.81 m/s.

    I've found the time it takes for the ball to drop using the equation

    v = vo + a t, t comes out to be 2.22s.

    But here on out, I don't know what to do next, i know that vx is constant and acceleration is zero in the x-direction, but I'm unable to get anything more. Any help will be appreciated, thanks in advance!
     
  2. jcsd
  3. Sep 27, 2007 #2
    lol I'm seeing a lot of projectile problems posted today.

    You've done the major part of calculation. The y-direction of velocity occurs when the projectile lands, meaning at that particular point there is no x-direction of velocity. And initially, when the ball is thrown you only have x direction of velocity. Does this give you a clue of how to solve the last part? You only have to do one simple arithmetic.
     
  4. Sep 27, 2007 #3
    OOPS OOPS, damn, forget my last response, that was misleading (Damn, I had too many beers last night). You DO have both x and y components of velocities when the ball lands. You've calculated the y-component of velocity and x-component of velocity stays the same, so your total velocity is

    [tex]v = \sqrt{v^2_x + v^2_y}[/tex]

    Then the problem says this is 4 times greater than the initial velocity, which only has x-component on it. Set the total velocity equal to [tex]\frac{v_x}{4}[/tex], solve for [tex]v_x[/tex]
     
    Last edited: Sep 27, 2007
  5. Sep 27, 2007 #4
    Oh I see, I finally got the answer, thanks a lot!
     
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