A ball is dropped from a height of 20m. When it bounces, it rebounds with a speed that is one half of the speed at which it hits the ground. How high does it bounce?
To find the height to which the ball bounces, we can use the concept of conservation of mechanical energy. The mechanical energy of the ball is conserved during its bounce, meaning the sum of its kinetic energy and potential energy remains constant.
Let's break down the problem step by step:
Step 1: Calculate the velocity of the ball just before it hits the ground.
The initial potential energy of the ball is converted into kinetic energy as it falls, neglecting any losses due to air resistance. The potential energy is given by the equation PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height from which it was dropped (20 m in this case).
Since the ball is falling from rest, the potential energy is equal to the kinetic energy just before it hits the ground. Therefore, we can write:
PE = KE
mgh = (1/2)mv^2
Simplifying the equation, we find:
gh = (1/2)v^2
v^2 = 2gh
v = sqrt(2gh)
Substituting the given values, we get:
v = sqrt(2 * 9.8 * 20) ≈ 19.8 m/s
Step 2: Calculate the velocity of the ball just after it bounces.
According to the problem, the ball rebounds with a speed that is half of the speed at which it hits the ground. Therefore:
v' = (1/2)v
v' = (1/2) * 19.8 ≈ 9.9 m/s
Step 3: Calculate the height to which the ball bounces.
Using the same equation (PE = KE), we can equate the potential energy at the maximum height to the kinetic energy just after bouncing.
(1/2)mv'^2 = mgh'
(1/2)(9.9)^2 = 9.8h'
Simplifying the equation gives:
h' = ((9.9)^2) / (2 * 9.8) ≈ 5.0 m
Therefore, the ball bounces to a height of approximately 5.0 meters.
Summary:
The ball bounces to a height of around 5.0 meters. To solve this problem, we used the principles of conservation of mechanical energy to relate the potential and kinetic energies of the ball before and after the bounce.