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Stone thrown from a building

  1. Jun 6, 2012 #1
    1. The problem statement, all variables and given/known data

    A stone is thrown upward from the top of a building at an angle of 30.0 degrees to the horizontal and with an initial speed of 20.0 m/s. The height of the building is 45.0 m.

    Where does the stone strike the ground?

    2. Relevant equations

    [itex]v_{x0} = v_0\cos{\theta_0}[/itex] where [itex]v_{x0}[/itex] is the original velocity x component, [itex]v_0[/itex] is the original velocity, and [itex]\theta_0[/itex] is the original angle (not sure if we use this)

    Also [itex]v_{y0}[/itex] is the original velocity y component.

    3. The attempt at a solution

    I'm not sure how to proceed here. We're finding horizontal distance, so no equations concerning [itex]v_{y0}[/itex] would work, and [itex]v_{x0} = v_0\cos{\theta_0}[/itex] only works if the vertical distance is the same as the original vertical distance..
     
  2. jcsd
  3. Jun 6, 2012 #2
    Hi professordad. If you're confident in using vectors and integrating, for this problem you can use [itex]\mathbf{F}=m\mathbf{\ddot{r}}=m(0,-g,0)[/itex], separate the components and integrate along with the boundary conditions [itex]\mathbf{x}_0=(0,h,0), \mathbf{v}_0=(v_0cos(\theta),v_0sin(\theta),0)[/itex]. After obtaining an equation for [itex]x(t)[/itex] and [itex]y(t)[/itex], substitute [itex]y=0[/itex] (when the y-axis meets the ground) and solve for t, and use this value in [itex]x(t)[/itex].
     
  4. Jun 6, 2012 #3

    berkeman

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    Staff: Mentor

    Doesn't seem like you need to use vectors or integration for this problem, unless I'm missing something.

    Just write the two equations for the motion in the vertical and horizontal directions. The horizontal velocity stays constant (ignoring air resistance). The vertical motion follows the equations for motion given a constant acceleration (gravity).

    You use the initial vertical velocity and position to figure out how long it takes the ball to hit the ground. Then use that time and the horizontal velocity to figure out how far out it hits.

    Lets see some equations!
     
  5. Jun 6, 2012 #4
    No not required, just a preference
     
  6. Jun 6, 2012 #5
    There are 2 equations of motion. Vertical and horizontal.
    They work independently
    They are only related by time since it is a single stone.
     
  7. Jun 6, 2012 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Using "vertical and horizontal directions" IS using vectors and the "equations for motion given a constant acceleration" are derived by integration. You can sweep them under the rug but they are there none the less.
     
  8. Jun 6, 2012 #7

    berkeman

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    Staff: Mentor

    True, but I was worried that mentioning that approach would scare the OP into thinking that the problem was more complicated than it is. :smile:
     
  9. Jun 7, 2012 #8
    First figure out how much time it takes for the stone to hit the ground using the fact that
    y=y0+vot+1/2gt2

    Knowing this time, you should be able to calculate the horizontal distance traveled, using the equation you posted
     
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