1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Horizontal projectile problem(revised)

  1. Sep 30, 2007 #1
    1. The problem statement, all variables and given/known data
    A person standing on a cliff throws a stone with a horizontal velocity of 15.0m/s and the stone hits the ground 47.0 m from the base of the cliff.

    A. How much time does it take for the stone to hit the ground?
    B. How high is the cliff?
    C. What was the impact velocity of the the stone(magnitude and derection)?

    you can ignore air resistance

    2. Relevant equations
    V = Vo + at
    X = Xo +Vot + 1/2at^2
    V^2 = Vo^2 + 2ax


    3. The attempt at a solution

    i drew out the diagram and set up my knowns but dont know which variable to look for first the time or the cliff hieght and how to manipulate the equations to do that
     
  2. jcsd
  3. Sep 30, 2007 #2
    A.) This is more simple than it appears. Since horizontal motion is independent of vertical motion, you can simply use d=vt. Isolate the x-direction. You have the object traveling 47 meters at 15 m/s. Find time.

    Let me know if you have trouble finding the other answers knowing this information...
     
  4. Sep 30, 2007 #3
    thanks alot i was making it much harder than it needed it to be, i will thank you
     
  5. Sep 30, 2007 #4
    wait how do i isolate x i just realized i didnt take very good notes...sry
     
  6. Sep 30, 2007 #5
    never mind
     
  7. Sep 30, 2007 #6
  8. Sep 30, 2007 #7
    yes sir i do after i get time which is 3.13 seconds i plug it into the equation id normally use to find time
     
  9. Sep 30, 2007 #8
    now how do i go about finding the derection in degrees at which the rock struck the ground? Sin(V)?
     
  10. Sep 30, 2007 #9
    The time that it taken for the rock to hit the ground is == the time allowed for the horizontal motion to move the rock in the x direction, from that you find the horizontal displacement.

    As for the angle it hits the ground at, that is a bit more complicated at the rock moves in an arc, which isnt subject to right angle triangle trigometric functions, the only way I would guess doing it would be to plot a graph and do it the practical way like that, however i'm waiting for PF mentor to blow me out the water with some awesome mathematical method of doing so (as I dont know the equation off the back of my head personally)
     
  11. Sep 30, 2007 #10
    Vectors. Vfinal will = sqrt(Vxfinal^2 + Vyfinal^2)
    and Vxfinal=Vxinitial, since horizontal velocity doesn't change from beginning to end.
    and Vyfinal= Vinitial*time - (1/2)(g)(t^2) I believe.

    But you asked for direction oops. Well that's how you do it anway. Direction will equal arctan(Vyfinal/Vxfinal).
     
  12. Sep 30, 2007 #11
    But it is! The mathematical method you mention is simply the inverse tangent of the final y velocity component divided by the final x velocity component. Its simple in vector terms.

    The triangle is of course formed by these final components of velocity. Vy final is the 'opposite' of your triangle and Vx is your 'adjacent'. I wish I could draw it for you, but hopefully that makes sense.
     
    Last edited: Sep 30, 2007
  13. Sep 30, 2007 #12
    hmm im pretty sure u are able to use Pathagorean's theory to do this
     
  14. Sep 30, 2007 #13
    thank you johnson im remembering that lecture that escaped me now thank you
     
  15. Sep 30, 2007 #14
    No problem
     
  16. Sep 30, 2007 #15
    Ah, i see, your using velocity vectors rather than distance ones ^_^ makes much more sense when put into that context, wouldnt the angle between it and the floor just be equal to:

    [tex]Tan(\theta)=V_{y}/V_{x}[/tex]

    so:

    [tex]Tan^{-1}(\frac{V_{y}{V_{x}})[/tex]

    EDIT: No idea why that bugged but i was using Tan = Opp/Adjacent

    or is [tex] ArcTan = Tan^{-1}[/tex]? if so, apologies, but i hate those inverse trigometric functions >,<
     
    Last edited: Sep 30, 2007
  17. Sep 30, 2007 #16
    Yes that's exactly it. Tan(theta)= Vy/Vx , which is the same as arctan(Vy/Vx)= theta. Yep!


    Unless I'm misusing my notation. I mean the inverse tangent when I say arctan, is that correct?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Horizontal projectile problem(revised)
  1. Horizontal projectile (Replies: 3)

Loading...