Projectile Motion of water balloon

In summary, the initial calculations for the water balloon's velocity components were correct. The vertical component is 10.5 m/s (up) and the horizontal component is 32.3 m/s (forward). To adjust the water cannon so that the balloon will hit the target, the angle of the cannon should be increased to a steeper angle. This can be found by solving the equation for the maximum horizontal range of the projectile, which is given by the formula R = (v^2 sin2θ)/g. By plugging in the values for velocity and gravity, we can solve for θ and find the angle that will give the maximum horizontal range and hit the target.
  • #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!
 
Physics news on Phys.org
  • #2
Evangeline101 said:

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
Evangeline101 said:

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
SteamKing said:
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)
 

Attachments

  • upload_2016-4-22_19-38-7.png
    upload_2016-4-22_19-38-7.png
    1.3 KB · Views: 389
  • #6
Evangeline101 said:
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:

SteamKing said:
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
Evangeline101 said:
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.

ehild said:
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
Evangeline101 said:
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
Evangeline101 said:
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?
 

1. What is projectile motion?

Projectile motion is the motion of an object through the air, where the only force acting on it is gravity. The object follows a curved path known as a parabola.

2. How does a water balloon follow projectile motion?

A water balloon follows projectile motion when it is thrown or launched into the air. The initial force propels the balloon forward, while gravity pulls it down, causing it to follow a curved path. As the balloon falls, its speed increases until it hits the ground or another object.

3. What factors affect the projectile motion of a water balloon?

The factors that affect the projectile motion of a water balloon include the initial velocity, the angle at which it is thrown, and the force of gravity. Other factors such as air resistance and wind can also affect the motion of the balloon.

4. How can we calculate the trajectory of a water balloon?

The trajectory of a water balloon can be calculated using the equations of motion and the values for initial velocity, angle, and gravity. These calculations can be done manually or with the use of a computer program. However, factors such as air resistance and wind can make the trajectory more difficult to predict.

5. How is projectile motion of a water balloon useful in real life?

The principles of projectile motion can be applied in many real-life situations, such as sports and engineering. For example, understanding the trajectory of a water balloon can help in designing water balloon launchers or predicting the path of a thrown object in a game of catch. It is also a fundamental concept in the study of physics and can help us understand the motion of other objects in the natural world.

Suggested for: Projectile Motion of water balloon

Replies
39
Views
2K
Replies
6
Views
973
Replies
20
Views
2K
Replies
18
Views
1K
Back
Top