A water balloon dropped from a dorm window does not break on impact. From what height would you need to drop the balloon to double the speed of the balloon?
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double the height.
Angel: You are wrong this time.
To determine the height from which you would need to drop the water balloon to double its speed on impact, we can use the principle of conservation of energy.
The potential energy of an object at a certain height is equal to its mass multiplied by the acceleration due to gravity (9.8 m/s^2) multiplied by the height:
Potential energy = mgh
Where:
m = mass of the water balloon
g = acceleration due to gravity
h = height from which the balloon is dropped
The kinetic energy of an object is given by the formula:
Kinetic energy = 1/2 * m * v^2
Where:
m = mass of the water balloon
v = velocity of the balloon at impact
Since we want to double the speed of the balloon, the final velocity will be twice the initial velocity. Let's represent the initial velocity as v1 and the final velocity as v2. So, v2 = 2 * v1.
According to the conservation of energy principle, the potential energy at the initial height should be equal to the kinetic energy at impact:
Potential energy = Kinetic energy
mgh = 1/2 * m * v2^2
mgh = 1/2 * m * (2 * v1)^2
mgh = 1/2 * m * 4 * v1^2
mgh = 2 * 1/2 * m * v1^2
mgh = m * v1^2
We can cancel out the mass (m) on both sides of the equation.
gh = v1^2
To double the speed (velocity) of the balloon, we need to solve for h:
h = v1^2 / g
Keep in mind that the height should be measured in meters, the velocity in meters per second (m/s), and the acceleration due to gravity as 9.8 m/s^2.
By plugging in the appropriate values for v1 and g, you can find the height needed to double the speed of the water balloon on impact.