Projectile Motion of water balloon

  • #1

Homework Statement


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.


Homework Equations

:[/B]
V1v= v1sinθ
V1h= v1cosθ
Δd = v1Δt + 1/2 a Δt2
Δdh = VhΔt

The Attempt at a Solution


a) [/B]
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!
 

Answers and Replies

  • #2
ehild
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Homework Statement


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.


Homework Equations

:[/B]
V1v= v1sinθ
V1h= v1cosθ
Δd = v1Δt + 1/2 a Δt2
Δdh = VhΔt

The Attempt at a Solution


a) [/B]
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!
How do you get shorter range of the projectile? How does the range depend on the initial speed and angle?
 
  • #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.
 
  • #4
SteamKing
Staff Emeritus
Science Advisor
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Homework Statement


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.


Homework Equations

:[/B]
V1v= v1sinθ
V1h= v1cosθ
Δd = v1Δt + 1/2 a Δt2
Δdh = VhΔt

The Attempt at a Solution


a)

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

This is the horizontal component of the initial velocity. What about the vertical component of the velocity?
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
Does gravity act in the horizontal direction? That's what you are saying with the equations above.
V1h = v1cosθ
V1h = 34.0 m/s cos18°
= 32.3 m/s (forward)
Whoops! You've gotten the components of the initial velocity swapped around.

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

View attachment 99550

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)
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.
 
  • #7
Oh okay. Ya I don't know, the font size changed when I posted the question :biggrin:

Correct your initial calculations before worrying about this.
So would you be able to help me figure out part b since the original calculations are correct?
 
  • #8
SteamKing
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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?
Sure. Just answer ehild's question in Post #2.
 
  • #9
Okay.

How do you get shorter range of the projectile? How does the range depend on the initial speed and angle?
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?
 
  • #10
SteamKing
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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?
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?
 
  • #11
ehild
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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?
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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