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Finding displacement

  1. Jul 15, 2014 #1
    Jamilla throws a stone horizontally off of a pier into the ocean at a velocity of 20 m/s. If the stone is 40.0 m 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?

    Vf= 20m/s
    Vi= 0m/s
    a=9.8m/s2
    Δd= 40m
    Δd=?

    Vf2=Vi2+2aΔd

    Vf=velocity final
    Vi= velocity initial
    a= acceleration
    Δd= displancement

    (rearrange the equation)
    Δd= (Vf2-Vi2) / 2a
    Δd= (20^2-o^2) / 2x9.8
    Δd= 400/19.6
    Δd= 20.4m

    ∴ the stone was 20.4m above the water surface when Jamilla threw it

    Not sure if i am doing this correctly?
    Much thanks in advance to those who help me out!
     
  2. jcsd
  3. Jul 15, 2014 #2
    Be careful - the initial velocity is horizontal, but there is no horizontal acceleration. You have that the initial vertical velocity component is 0 m/s, the vertical acceleration is 9.8 m/s^2 downwards, and you don't know the final vertical velocity. You know the horizontal distance travelled is 40m, and the horizontal speed is constant at 20 m/s since there's no horizontal acceleration, so you can find the time of flight using speed = distance / time.

    Then you'll have enough information to calculate the height using the vertical information...
     
  4. Jul 16, 2014 #3

    CWatters

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    What Naz93 said. Break it down into horizontal and vertical components and solve in two stages. You appear to have them a bit mixed up.
     
  5. Jul 16, 2014 #4
    You are approaching the wrong way. The stone travels both in horizontal as well as vertical direction. And motion in two different directions has nothing to do with each other. Here, the acceleration due to gravity acts only in the y-direction. So, the stone accelerates downward. On the other hand, there is no acceleration in x-direction. Hence, the stone undergoes uniform motion in x-direction. It is called a projectile motion:-

    x= v(i,x)* time v(i,x): initial velocity in x-direction
    y= 1/2* g * time^2 g:acceleration due to gravity note: v(i,y)=0

    Replacing time from both the equations:-

    y=(g * x^2)/(2*(v(i,x))^2)

    Putting values:-

    y=(10* 40^2)/(2* 20^2)= 20 m
     
    Last edited: Jul 16, 2014
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