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Projectile Motion - Rock Thrown Horizontally Off a Cliff

  1. Sep 16, 2011 #1
    1. The problem statement, all variables and given/known data
    A student stands on the edge of a cliff and throws a stone horizontally over the edge with a speed "v1". The cliff is "h" meters high. Given [h,v1], Determine:
    a. The time to hit the ground
    b. The horizontal distance traveled
    c. The magnitude and direction of the stone's velocity just before hitting the ground


    2. Relevant equations
    x direction
    [tex]
    V_x=V_{0x}
    [/tex]
    [tex]
    x=V_0\cos{\theta}
    [/tex]

    y direction
    [tex]
    V_y=V_{0y}-gt
    [/tex]
    [tex]
    y=V_{0y}t-\frac{1}{2}gt^2
    [/tex]
    [tex]
    V_y^2=V_{0y}^2-2gy
    [/tex]

    3. The attempt at a solution
    If I set up my coordinate system so that 0 is ground level, and h is cliff level (where the stone was thrown from), and the distance to the landing point of the stone when it hits the ground is, say "r" meters from 0 in the x direction.

    I've shown what work I have in the attached images, but to be perfectly honest, I'm having trouble with the entire basic strategy to approaching this problem.

    When I try to solve for x distance r, I need time. So when I try to solve for time, that becomes reliant on r. I can't even think of anything else I can find.

    I feel I'm either not utilizing the possible angles within this problem, or I've completely missed some fundamental idea in regards to projectile motion.

    A good shove in the right direction on this kind of problem would be VERY much appreciated.
     

    Attached Files:

  2. jcsd
  3. Sep 16, 2011 #2
    Here's a simple approach:-

    Firstly make a table of Horizontal motion and vertical motion, like according to your question,
    [​IMG]

    H-Horizontal motion
    V-Vertical motion
    u-initial velocity
    s-displacement
    h-Height of the cliff
    x and t are assumed since we have to find them out.

    Ok, first lets see in which column, assumed variable are less. Its obvious that in V column, there is only one variable "t". So can you apply equation of motion here? :smile:
     
  4. Sep 16, 2011 #3
    Something tells me I've once again forgotten a rather important fact about the equation
    [tex]y=V_{0y}t+\frac{at^2}{2}[/tex]

    Is it correctly written as this by chance :grumpy:
    [tex](y-y_0)=V_{0y}t+\frac{at^2}{2}[/tex]
     
  5. Sep 16, 2011 #4
    You can take y0 as 0 since the initial displacement is zero. :smile:

    Now just plug in the values and you will get the time. :)
     
  6. Sep 16, 2011 #5
    How about this then:
    [tex](y-y_0)=V_{0y}t+\frac{at^2}{2}[/tex]
    [tex](0-h)=0-\frac{gt^2}{2}[/tex]
    [tex]2h=gt^2[/tex]
    [tex]\frac{2h}{g}=t^2[/tex]
    [tex]\sqrt{\frac{2h}{g}}=t[/tex]
     
  7. Sep 16, 2011 #6
    That's right. :smile:

    Now just see that we have t same for both H and V motion. So now we are left with x in H motion. Try applying the suitable equation of motion in the H column.
     
  8. Sep 16, 2011 #7
    Here is my attempt for Horizontal distance covered:
    [tex]x=(V_0\cos{\theta})t[/tex]
    [tex]x=V_1t[/tex]
    [tex]x=\frac{V_1\sqrt{2h}}{g}[/tex]
    hows that look?

    Now in order to get the direction and magnitude of the end velocity of the rock, I need both the Vx and Vy right?
     
  9. Sep 16, 2011 #8
    Yep, that's right. :smile:

    Now you have solved the first two parts. About the third part, you can go like this:-
    Find the final velocity in Horizontal and Vertical direction, then find their resultant. :)
     
  10. Sep 16, 2011 #9
    Alright, so far so good, lets see if I got this right:

    To find the Velocity in the y direction near point of impact
    [tex]V_y=V_{0y}^2+2a(y-y_0)[/tex]

    substitute in my knowns:
    [tex]V_y=(0)-g(\frac{\sqrt{2h}}{g})[/tex]
    [tex]V_y^2=(0)-2g(0-h)[/tex]
    [tex]V_y^2=-2g(-h)[/tex]
    [tex]V_y^2=2gh[/tex]
    [tex]V_y = \sqrt{2gh}[/tex]

    As for the resultant, are you referring to vector addition?

    If so, I think I got [tex]<V_1,0,0>+<0,\sqrt{2gh},0>\ = \ <V_1,\sqrt{2gh},0>[/tex]

    Hows all of this look?

    But what if the question asked for the angle, would I need to use the Tangent Function and take the Inverse Tangent to find Theta?
     
  11. Sep 16, 2011 #10
    You got the final velocity right but with a long method. :smile:
    What i expected was that you would have used the first equation of motion.

    Yep, i am referring to vector addition. I think you did it by co-ordinates. I have never used that way. :smile:
    We have horizontal and vertical final velocities. These both are perpendicular to each other.
    So [tex]V=\sqrt{V_x^2+V_y^2}[/tex]

    V is the resultant velocity.
    Vx=v1
    Vy=[itex]\sqrt{2gh}[/itex]

    Just plug in the values and your are done. :smile:

    If you want to find the angle, you need to first specify the reference, Horizontal or vertical. :)
     
  12. Sep 16, 2011 #11
    Well, it seems I got a grasp of the problem now, I think the only thing that messed me up in the beginning was the fact that I was applying one of the formulas wrong.

    Thanks for all the help, its much appreciated :smile:
     
  13. Sep 16, 2011 #12
    Your welcome!!:smile:
    And remember the table approach. Whenever you deal with projectiles, the best way is to make the table :smile:

    I forgot to welcome you. Welcome to the board. Hope you enjoy your stay here. :smile:
    Before wandering on the board, have a look at the rules. They may help you.
     
  14. Sep 17, 2011 #13
    Thanks for the advice

    Thanks again
     
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