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Projectile Motion of water balloon

  1. Apr 21, 2016 #1
    1. The problem statement, all variables and given/known data
    A water balloon is fired at 34.0 m/s from a water cannon, which is aimed at an angle of 18° above the ground. The centre of the cannon's target (which has a radius of 1.0m) is painted on the asphalt 42.0 m away from the water cannon. Will the balloon hit the target? Justify your response with calculations that indicate where the water balloon will land and make one suggestion about how to adjust the water cannon so that the water balloon will hit the target. Justify your choice.


    2. Relevant equations:
    V1v= v1sinθ
    V1h= v1cosθ
    Δd = v1Δt + 1/2 a Δt2
    Δdh = VhΔt

    3. The attempt at a solution
    a)

    V1v = v1sinθ
    V1v = 34.0 m/s sin18°
    = 10.5 m/s

    Δd = v1Δt + 1/2 a Δt2
    0= (10.5 m/s) Δt + 1/2 (9.8 m/s2) Δt2
    0 = Δt (10.5 m/s -4.9 m/s2 Δt)
    Δt = 2.1s

    V1h = v1cosθ
    V1h = 34.0 m/s cos18°
    = 32.3 m/s (forward)

    Δdh = Vh Δt
    Δdh= (32.3 m/s [forward]) × (2.1s)
    Δdh = 68 m [forward]

    Therefore the ball will not hit the target but pass it by 26 m. Is this correct?

    b) For this part, I know I have to adjust the angle of the water cannon but I am not sure how to find out what angle I should use.

    Any help would be greatly appreciated!
     
  2. jcsd
  3. Apr 21, 2016 #2

    ehild

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    How do you get shorter range of the projectile? How does the range depend on the initial speed and angle?
     
  4. Apr 21, 2016 #3
    Find the permissible values for the "Maximum Horizontal Range of the projectile".
    Use the formula for "Maximum Horizontal Range of the projectile" and solve it for Θ.
     
  5. Apr 21, 2016 #4

    SteamKing

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    This is the horizontal component of the initial velocity. What about the vertical component of the velocity?
    Does gravity act in the horizontal direction? That's what you are saying with the equations above.
    Whoops! You've gotten the components of the initial velocity swapped around.

    This is the missing vertical component of the initial velocity!
    You need to correct the components of the initial velocity of the water balloon.
    Correct your initial calculations before worrying about this.
     
  6. Apr 22, 2016 #5
    How are they swapped?
    This is what it explains in my lesson: vertical component being the first equation and horizontal being the second:

    upload_2016-4-22_19-39-16.png

    So for vertical component of initial velocity:
    V1v = v1sinθ

    V1v = (34.0 m/s) sin18° (up)
    = 10.5 m/s (up)

    For horizontal component of initial velocity:

    V1h = v1 cosθ

    V1h = (34.0 m/s) cos 18° (forward)
    = 32.3 m/s (forward)
     

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  7. Apr 22, 2016 #6

    SteamKing

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    Your original calculations are correct. I got distracted because the weird formatting made them look like an eye chart - big letters starting out, and smaller letters the further down you read.
     
  8. Apr 22, 2016 #7
    Oh okay. Ya I don't know, the font size changed when I posted the question :biggrin:

    So would you be able to help me figure out part b since the original calculations are correct?
     
  9. Apr 22, 2016 #8

    SteamKing

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    Sure. Just answer ehild's question in Post #2.
     
  10. Apr 22, 2016 #9
    Okay.

    To get a shorter range of the projectile, you would set the launch angle to an angle higher than 45° because horizontal displacement decreases at angles higher than 45°. As the initial velocity increases, the vertical height and horizontal displacement also increase.

    Is this correct?
     
  11. Apr 22, 2016 #10

    SteamKing

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    Is that all? What about the launch speed?

    As far as the angle of launch is concerned, 45° gives you the theoretical maximum range of a projectile without accounting for drag. However, this balloon was launched at an angle of 18°. Would you necessarily need to increase the launch angle above 45° to hit this target, assuming the launch speed stays the same?
     
  12. Apr 22, 2016 #11

    ehild

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    You got too long range even at angle of 18 degrees. And the maximum range is at 45˘. Are you sure that you need to set the launch angle higher than 45°? What happens when you decrease the launch angle?
    And what about changing the initial speed?
     
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