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?
V^2 is proportional to height.
V is proportional to sqrt(Height)
If sqrtH doubles, H must increase by a factor of 4.
To determine the height from which the water balloon needs to be dropped to double its speed upon impact, we can analyze the concept of gravitational potential energy and kinetic energy.
When an object falls freely under the influence of gravity, its potential energy is converted into kinetic energy. Assuming there is no significant air resistance, we can equate the initial potential energy to the final kinetic energy after the fall:
Potential Energy = Kinetic Energy
The potential energy is given by the equation:
Potential Energy = m * g * h
Where:
m = mass of the object
g = acceleration due to gravity (approximately 9.8 m/s²)
h = height from which the object is dropped
The kinetic energy is given by the equation:
Kinetic Energy = (½) * m * v²
Where:
v = velocity of the object
Since we want to double the speed of the balloon, the final velocity will be twice the initial velocity. Therefore, we can express the final kinetic energy as:
(½) * m * (2v)² = (½) * m * 4v² = 2 * (½) * m * v²
Now, we can equate the potential energy and the final kinetic energy:
m * g * h = 2 * (½) * m * v²
The mass of the balloon cancels out on both sides of the equation, simplifying it to:
g * h = 2 * v²
Since the initial velocity is zero (the balloon was dropped), the equation further simplifies to:
g * h = 2 * (0²) = 0
This means that no matter the height from which the balloon is dropped, it will always have a speed of zero upon impact.
Therefore, there is no specific height from which the balloon needs to be dropped to double its speed upon impact because it will always be zero.