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!

Physics Vectors

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data
    I have this question regarding how i would do the second part of this question as i already know how to do the first part ,but i can't seem to find an approach to the second."What is the speed of the stone at a height of 10m from the ground?"

    A stone is thrown off the edge of a cliff and lands 37.5m from the base of the cliff.If the stone's initial speed was 11.5m/s,how high is the cliff?
    What is the speed of the stone at a height of 10m from the ground?


    2. Relevant equations



    3. The attempt at a solution
    tx=dx/vx
    37.5/11.5m/s =3.26 secs

    y=1/2gty^2
    y=.5(9.8m/s^2)(3.26s)^2 = 52 m
     
  2. jcsd
  3. Dec 6, 2011 #2
    Firstly, the question should have states "how" the stone was launched. I will assume it was horizatontally thrown with the given speed. If this is the case, then you must split the problem into two orthogonal directions, the obvious choice being x,y (where y is vertical).

    Then you know the distance and velocity in the x direction and have correctly identified the time between it being launched and it striking the ground:
    you have then used
    [itex] y=\frac{1}{2} g t_{y}^{2} [/itex]
    but for simplicity we can remember that tx=ty otherwise the situation would be very strange, and so
    y=53m

    What is the speed of the stone at a height of 10m from the ground?

    Okay so now you need a function which will give you the velocity of the stone as a function of some other things, since you need not include any relevant equations, namely the equations of motion http://http://en.wikipedia.org/wiki/Equations_of_motion I think you will find all you looking for in equation 5

    [itex] v^{2}=u^{2}+2as [/itex]
     
  4. Dec 6, 2011 #3
    Ok so using the equation you gave me i subtracted the height 52m -10m from the ground resulting as 42.1m.

    v2^2=11.5m/s^2+2(9.8)(42.1m)

    v2=30.94 m/s Would this be correct or would i have to a use the range equation for any reason R=x^2+y^2?
     
  5. Dec 7, 2011 #4
    I am unsure what you mean by the range equation, but from it's form I would guess you mean
    r^2=x^2+y^2 , that is Pythag' which would be the square of the physical distance the rock lands away.

    You are asked for the "speed" which I am sure you are aware is a magnitude of the velocity (a vector). So you have calculated the y component of the velocity, and you know the x component (since gx=0) therefore you must calculate the magnitude of the velocity.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Physics Vectors
  1. Vectors in Physics (Replies: 1)

  2. Vectors in physics (Replies: 2)

  3. Vectors and Physics (Replies: 8)

  4. Physics- vectors (Replies: 1)

Loading...