# Water balloon lab

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1. Oct 31, 2016

### AStupidHippo

1. The problem statement, all variables and given/known data
"In part 2 of the lab you will be firing a water balloon over a tall fence, aiming to hit a target." The fence is 3 meters high. You may fire the balloon from anywhere behind the fence, up to 4 meters away from it: the target will be 20 meters away. Find the angle to shoot the balloon and the stretching distance. There are multiple correct answers.
horizontal distance=s = 20 to 24 meters
mass= .25kg
Spring constant= k= 336.4
spring stretching distance= x = 0 to 25 centimeters
theta=?
2. Relevant equations
PE + W = KE
PE=.5k x^2
w = Fk * s
Fk = k*x
KE= .5mv^2
3. The attempt at a solution

Assuming we use all 24 meters
.5(336.4)(x^2) + 24(336.4)(x) = .5(.25)v^2
v= sqrt(2691.2((x^2)/2)+24x)

We're lost after this. Even plugging in a value for x gives us a very large velocity.
For example plugging x=.2 gives a velocity of ~113 m/s. This doesn't seem realistic.

2. Oct 31, 2016

### TomHart

Sorry I don't have a lot of time. My ride will be here in minutes.

If I was going to work this problem, I would be wondering how much energy would be needed to get the balloon over the fence and to the target. Obviously, if I am extremely close to the fence when I launch, the balloon will have to go extremely high. You may not have that much energy. All of that energy has to come from the launching device. How high is the balloon when it is launched? Can you place the launching device as high as you want or does it have to be at ground level? (I assume it is the latter.) What is the height of the balloon when it is released from the launching device? Will the launching device provide enough energy at the horizontal position I selected. (I think 24 meters may be a good choice.)

I think you ought to take a second look at the equation you wrote.