A ball is dropped from the initial height h1 above the concrete floor and rebounds to a height of equal to 60% of the initial height h1. What is the ratio of the speed of the ball right after the bounce to the speed of the ball just before the bounce? This ratio is called coefficient of restitution e=v2/v1.
Presumably by the time you get to physics you can convert a % to a decimal or a fraction.
Just do it. That's the answer!
i think this is a pre-calculus question
To find the ratio of the speed of the ball right after the bounce to the speed of the ball just before the bounce (coefficient of restitution), we can use the principle of conservation of mechanical energy.
Let's assume that the mass of the ball is m and the initial height is h1. The potential energy at the initial height is given by mgh1, where g is the acceleration due to gravity. Initially, the ball only has potential energy, so its initial kinetic energy is zero.
As the ball drops to the concrete floor, it converts all of its potential energy into kinetic energy. So at the point of impact, the kinetic energy is equal to the potential energy, i.e., 0.5mv1^2 = mgh1.
After the ball rebounds, it reaches a height equal to 60% of the initial height, which is 0.6h1. At this point, the ball only has potential energy, and its kinetic energy is zero again.
Using the principle of conservation of mechanical energy, we can equate the potential energy at the maximum height to the initial kinetic energy. Therefore, 0.5mv2^2 = mgh1 * 0.6.
Let's simplify the equations:
0.5mv1^2 = mgh1 --> v1^2 = 2gh1
0.5mv2^2 = mgh1 * 0.6 --> v2^2 = 2gh1 * 0.6
Now, divide both equations:
v2^2 / v1^2 = (2gh1 * 0.6) / (2gh1) = 0.6
Taking the square root of both sides:
v2 / v1 = √0.6
Therefore, the ratio of the speed of the ball right after the bounce to the speed of the ball just before the bounce (coefficient of restitution) is √0.6.