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Velocities with a resulting horizontal component

  1. Nov 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Romeo is chucking pebbles gently up to Juliet's window, and he wants the pebbles to hit the window with a horizontal component of velocity. He is standing at the edge of a rose garden 8m below her window and 9m from the base of the wall. How fast are the pebbles going when they hit her window.


    2. Relevant equations
    d=.5at^2
    t=d/v
    t=√2d/a

    3. The attempt at a solution
    I don't know how to find the speed when it has only a horizontal component of speed. Am I supposed to find how fast he should throw it so its apex is at her window?
     
  2. jcsd
  3. Nov 22, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    A velocity has both vertical and horizontal components of velocity but the acceleration due to gravity applies only to the vertical component. Determine what the initial vertical velocity must be so that it is 0 at a height of 8m. Use that to determine the time it takes the pebble to go that high. Determine the horizontal speed speed so that the pebble will go 9 m in that time.
     
  4. Nov 22, 2012 #3
    Yes. At the apex of the arc, the derivative of the vertical displacement with respect to the horizontal displacement is 0. In non-mathy terms, this means that the pebble is moving horizontally at the apex of the arc.
     
  5. Nov 22, 2012 #4
    The main idea is to remember that you can decompose this type of motion (known as projectile motion) into two parts; the motion in the vertical direction and the motion in the horizontal direction. If you take the origin of your coordinate system to be ground (assuming falsely that Romeo has zero height, or is otherwise in a hole in the ground such that he can throw the pebble at ground level) you will find that the equation of motion in the vertical component will be:
    $$ y(t) = \frac{-1}{2}gt^2 + V_0\sin(\theta)t $$

    Where [itex]V_0\sin(\theta)[/itex] is the vertical component of the initial velocity. The horizontal motion is given by:
    $$ x(t) = V_0\cos(\theta)t $$
    Where [itex]V_0\cos(\theta)[/itex] is the horizontal component of the initial velocity. Now, we want that the vertical component of the velocity to be zero when the height above the ground is 8 m. From the vertical motion equation we can find the general expresion for the vertical velocity:
    $$ V_y(t) = -gt + V_0\sin(\theta) $$
    In order to find the time it takes the pebble to reach its apex (where its vertical velocity is zero) set the previous equation to zero and solve for the time t. Substitute this value into the motion equation and set it equal to 8 (Since at that time the height of the pebble will be 8). This will allow you to solve for an expression involving the initial velocity and the initial angle [itex]\theta[/itex]. Now, you know that the horizontal motion will be 9 m, so use that to solve for [itex] V_0 [/itex] and [itex]\theta[/itex]. I'll be happy to provide any more detail as needed.
     
    Last edited: Nov 22, 2012
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