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Two Dimensional Motion

  1. Jul 13, 2014 #1
    Problem:
    Jamilla throws a stone horizontally off a pier into the ocean at a velocity of 20 m/s. If the stone is 40.0m from the edge of the pier when it hits the water, how high above the water's surface was the stone when Jamilla threw it?

    Related Equations:
    displacement = velocity x time
    displacement = (v1) (t) + (a) (t2) /2
    t= square root of (2) (d) / a


    Solution:
    t= square root of (2) (d) /a
    t= square root of (2) (40.0m) /9.8m/s2
    t= square root of 80m/ 9.8m/s2
    t= square root of 8.16
    t= 2.9s

    d= (v)(t)
    d= (20m/s) (2.9s)
    d= 58.0m

    t= time
    d= displacement
    a= acceleration

    Having trouble with the equations again. I want to know whether the equation I'm using is right or do I have to use a different equation?
    Therefore the stone was 58.0m high above the ground when Jamilla threw it.
     
  2. jcsd
  3. Jul 13, 2014 #2
    Here is also the same problem ,the given velocity of stone is velocity of stone in x direction.Initial Velocity in y direction is zero.Reply if you need help.
     
  4. Jul 13, 2014 #3
    Reply to Satvik Pandey

    Sorry but I'm not really understanding. Am I supposed to substitute velocity as 0 instead of 20m/s?
     
  5. Jul 14, 2014 #4

    Bandersnatch

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    Gold Member

    Hi kathy,

    It's looks like you're using the equations willy-nilly. It's a bad approach, as you just crunch the numbers instead of understanding what's going on.
    Let's take a step back and look at how equations of motion work.

    There is really just one equation, that looks like this:
    ##r=r_0+V_0t+\frac{1}{2}at^2##
    r is the final position(displacement), r0 is the initial position, and V0 is the initial velocity(everything with a subscript "0" means initial), and a is the acceleration. All these values are vectors(don't worry if you haven't studied them yet). t stands of course for time.

    Now, for 2-dimensional motion, usually it's the easiest to write the above equation separately for the horizontal(x) and vertical(y) directions:
    ##x=x_0+V_{0x}t+\frac{1}{2}a_xt^2##
    ##y=y_0+V_{0y}t+\frac{1}{2}a_yt^2##
    In 3-d motion you'd add another equation for z.
    Notice that the only thing that connects these equations is time t. All the other values are independent, and you can't substitute one for any other.

    Anyway, x0 now means how far in the horizontal direction the motion starts. For example, somebody throws a stone while standing 20 m away from point 0. y0 means how high the motion starts.
    V0x and V0y mean how fast the thing is moving when the motion starts - in horizontal or vertical direction. For example, throwing a stone directly upwards would mean it has got some non-zero velocity V0y.
    ax and ay are accelerations in either direction. When dealing with throwing stones and other ballistic motion, it's usually just one acceleration downwards, due to the force of gravity.
    Every time you want to indicate direction, you use positive values(+ sign) for right and up, and negative(- sign) for left and down.

    What you do, is you rewrite the two equations for x and y, deleting those parts that come out to 0.

    For example, imagine you're standing on top of a building, and just drop a stone from your hand without throwing it.
    If there is no acceleration in the horizontal direction(a=0), and the motion starts at point 0(x0=0), and the horizontal velocity is zero(V0x=0), the first equation reduces to ##x=0+0t+\frac{1}{2}0t^2=0##, which means that there is no motion in the horizontal direction.
    In the vertical direction, you get y0 initial displacement(the building is some height above ground), 0 initial velocity(no throwing), and -g acceleration(gravity pulls the stone down. Down means the minus sign). So you end up with ##y=y_0+0t-\frac{1}{2}gt^2##



    Now, back to your problem. Can you rewrite the two eqations for x and y:
    ##x=x_0+V_{0x}t+\frac{1}{2}a_xt^2##
    ##y=y_0+V_{0y}t+\frac{1}{2}a_yt^2##
    as I've shown you, only using what you know about the motion from the problem description? Think which values are zero and which are not.

    Once you've got that sorted out, it's just a matter of doing some algebra.
     
    Last edited: Jul 14, 2014
  6. Jul 14, 2014 #5
    Yes for calculating time required for stone to reach the ground substitute u(initial velocity in y direction)=0 and then multiply this 't' by 20 to get the answer.By the way y direction means vertical direction and x direction means horizontal direction.Reply if you need help.
     
    Last edited: Jul 14, 2014
  7. Jul 16, 2014 #6

    CWatters

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    Homework Helper

    With most problems of this type you can divide the information provided in the problem statement into horizontal and vertical components. You can then write one set of equations that address the horizontal motion and another set that address the vertical motion. This gives you a set of simultaneous equations that can be solved. The initial post lists a set of equations without stating if they apply to the horizontal or vertical components of motion. Best start by doing that.

    Sometimes it may appear that you are one parameter short of the number needed to solve the equations.. but remember that the time of flight is (usually) the same in both the horizontal and vertical planes.
     
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