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Fish shoots water droplet at bug

  1. Oct 6, 2007 #1

    ~christina~

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    1. The problem statement, all variables and given/known data
    A Sharp Shooter fish is a fish that hunts insect from the water. It is able to project water droplets toward insects causing them to drop into the water where they can be eaten. In the figure below the sharp shooter is at rest and spots a insect at rest on a leaf hanging over the water. The sharp shooter is 2.2 m away from the bug when it fires a droplet with an initial velocity of 5.98m/s at an agle of 65 degrees relative to the surface of the water. The droplet strikes the bug causing it to drop vertically downward onto the surface of the water.

    a) find the maximum height that the water droplet reaches and the time required to reach this height

    b) how long will it take the droplet to reach the bug ater leaving the fish's mouth? (ignore air resistance)

    c) the fish wants to claim it's prey immediately so that no other fish can steal it's meal. What must the minimum acceleration of the fish be if it is to arrive at the point directly beneath the bug at the same time the bug strikes the surface of the water?

    d) sketch graphs of the sharp shooter's displacement, velocity, and acceleration vs time for the time interval required for the fish to claim it's prey.

    [​IMG]
    2. Relevant equations
    kinematic equations?? I don't know which ones to use...

    for the first part...
    h= [vi^2 sin^2(theta)]/ 2g

    t= (vi sin (theta))/ g



    3. The attempt at a solution

    I know that :
    vi= 5.98m/s
    theta= 65 deg
    dx= 2.2m
    g= 9.8m/s^2


    a) max height reached and time it takes
    h= [vi^2 sin^2(theta)]/ 2g
    h= [(5.98m/s)^2 * sin^2(65)]/ 2*(9.8m/s^2) = 1.498=> 1.50m

    t= (vi sin (theta))/ g
    t= [(5.98m/s)*sin(65)]/9.8m/s^2 = .553sec

    b) I have no idea how to get this

    c)I don't know about this either

    d) I can't do that unless I find the previous either


    I don't know which kinematic equation to use to find the distance..how would I incorperate the angle into the equation if I need to...which I'm not sure about that either.

    Thanks
     
    Last edited: Oct 6, 2007
  2. jcsd
  3. Oct 6, 2007 #2

    Astronuc

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    Staff: Mentor

    Last edited: Oct 6, 2007
  4. Oct 7, 2007 #3

    ~christina~

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    For the first part I found from the application where you could plug it in that I had gotten it alright but I'm confused about the second part still.

    b.) for the application where you could plug it in it is for when you want to calculate the time of flight of the object.

    The reason I'm confused is because when you use the distance of the x direction..dx
    doesn't that according to the applet mean the whole flight distance including after the net in the applet? which would then make the distance if I use 2.2m incorrect since that would if I apply it to the applet picture, would be the equivalent of the distance to the net + the back of the net.


    well assuming that when they mean the distance to the net/ in this case the leaf with the bug on it...

    I plugged in the initial launch velocity of the droplet (vo= 5.98m/s), horizontal range (dx= 2.2m), and the angle (theta= 65 deg)
    and the applet came up with:

    time of flight as = 0.8705 s

    the height was calculated as d= 1.004m
    (this I think means that the point the object will pass through is 1.004m high which would mean that it hits the bug)

    ~What I have a problem with is how they found the time..I just can't figure out how they got the answer of t= 0.8705 s from the equations they have there...

    I DID find the time however using one of the equations for x= Vox*t
    I got the time by getting Vox first:

    Cos theta= Vox/Voy
    (Cos= adj/hyp)
    then plugging in..
    Cos 65= Vox/5.98
    5.98m/s Cos 65= Vox

    Vox= .8695m/s (this is close to the .870 that the applet calculated)


    c.) for this part where I have to find the acceleration of the fish if it ends up right below the bug when the bug hits the water.

    I think I have to find the time that it takes for the bug to hit the water and use that to plug into the equation to find the acceration for the fish..

    using y= 1/2gt^2
    the distance for the y direction of the bug is 1m but it is actually -1m since assuming the original distance is where the bug is and the area below is the negative direction it would also be negative...

    -1m= 1/2 (-9.8m/s^2)t^2
    -1m/-4.9m/s^2= t^2

    t= .4518s (time it takes for the bug to hit the water)

    To find the acceleration of the fish to reach the bug..
    dx= 2.2m
    t= .4518

    Then I'm not sure what do I do to find the accleration of the fish since I have no initial velcocity of the fish...or is it 0?

    How would I find this??

    d.) which equations would I use for graphing the displacement, velocity (I don't even know which equation for that since I didn't need to find that and I currently don't know how to find that), and accleration...I need to find that..

    would the displacement vs time graph be from the the equation..

    y= 1/2gt^2 ?


    Thanks Astronuc:smile:
     
  5. Oct 7, 2007 #4

    Astronuc

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    For part c, assume the fish is initial at rest, i.e. v = 0, and then it accelerates with constant acceleration.


    for part d, one simply needs to find x(t), v(t), a(t) and plot them as a functions of time. Now if a(t) = a (constant), what will that look like?
     
  6. Oct 7, 2007 #5

    ~christina~

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    Well since you didn't say anything about the first part I'll assume that it was correct...

    For c.) to find the acceleration of the fish so that it reaches the bug when it hits the water...

    I used the information I got before for the time the bug hits the water and ...

    then taking from that that the time is
    t= .45s
    ax=?
    vix= 0
    dx= 2.2m

    x= vox*t + 1/2*at^2

    2.2m= 0(.45s) + 1/2(a)(.45)^2

    2.2m= .10125a

    a= 21.73m/s^2

    d) for d answering what you asked...if accleration is constant wouldn't that look like a straight line?

    distance vs time:
    I can't really figure out the equation
    since..
    xf= xi + vxi*t + 1/2*ax*t^2

    but

    xi= 0
    vxi= 0 so wouldn't the equation only involve the accleration and the variable of time ?

    acceleration vs time:
    acceleration is constant? I assume it is since you said it was constant but how would I know that from what I have?

    I did find that accleration for the fish though but I didn't know if it was constant or not.

    a(t)= 21.73m/s^2

    velocity vs time

    vxf= vxi + ax*t

    vxi= 0
    ax= 21.73m/s^2

    v(t)= 21.73m/s^2*t ==> is this fine for the equation of the velocity vs time ?


    Thank You Astronuc =D
     
  7. Oct 8, 2007 #6

    ~christina~

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    distance vs time

    xf= xi + vxi*t + 1/2*ax*t^2

    I went ahead since you didn't reply...

    Xi=0
    Vxi= 0
    ax= 21.73m/s^2

    xf= 1/2 ax (t)^2
    x(t)= 1/2(21.73m/s^2)t^2

    x(t)= 10.865m/s^2 (t)^2
     
  8. Oct 9, 2007 #7

    Astronuc

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    The plot of constant acceleration would be a horizontal line with acceleration as the ordinate and time as the abscissa.

    The v (t) = a t assuming v0 = 0.

    Then integrating again would give a quadratic function, which if measured from the fishes initial position would have x(0) = 0, and x(t) = 1/2 a t2, which one has done correctly.


    Now if the reference for distance (position) was the location under the fish, then one would write

    x(t) = 2.2 - 10.865 t2 m

    and v(t) = - 21.73 t (m/s), since the fish is headed in the -x direction, and acceleration 'a' would = - 21.73m/s2.
     
  9. Oct 9, 2007 #8

    ~christina~

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    Thanks Astronuc for all your help :smile:
     
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