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?
A man is standing on a cliff and throws a ball outward 50m/s at an angle of 53 degrees. It hits the base of the cliff in 10 seconds.
Find:
a) The height of the cliff
b) the angle at which the ball goes into the sea
c) if the ball will hit a boat 150m out to sea.
h(t) = h - g/2to^2
After you drop the ball from a height of 20 feet and it hits the ground, it would reach a height of 15 ft
(just multiply your h by 3/4 each time to get the heights from each time it bounces...
20 * (3/4) = 15ft
To determine how high the ball bounces, we need to take into account the conservation of energy.
Let's break down the problem into different parts:
1. Initial drop: The ball is dropped from a height of 20 meters, so it has a potential energy equal to mgh, where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height (20 meters in this case).
2. Just before impact: The ball reaches the ground, and all its potential energy is converted into kinetic energy as it falls. The kinetic energy is given by (1/2)mv², where v is the velocity of the ball just before impact.
3. Bounce: After the ball hits the ground, it rebounds with a speed that is one-half of the speed at which it hits the ground again. Let's call this rebound velocity v'.
Now, to calculate how high the ball bounces, we need to find the height it reaches on the rebound.
The total energy just before impact is equal to the total energy just after the rebound. Therefore, we can equate the kinetic energy just before impact with the potential energy at the rebound height.
(1/2)mv² = mgh'
Where h' is the height of the rebound.
Since the ball rebounds with a speed that is one-half of the speed at which it hits the ground, we have v' = (1/2)v.
Substituting v' into the equation above:
(1/2)mv² = mgh'
We can cancel the mass (m) on both sides of the equation:
(1/2)v² = gh'
Now, we can solve for h':
h' = (1/2)v²/g
However, we need to determine the value of v, the velocity just before impact. Since the ball is dropped, it falls freely under the influence of gravity. Using the equation for final velocity (v) in free fall starting from rest:
v = √(2gh)
Substituting this value of v in the equation for h', we get:
h' = (1/2)[√(2gh)]²/g
Simplifying further:
h' = (1/2)(2gh)/g
The g cancels out:
h' = h
Therefore, the height to which the ball bounces is equal to the original height from which it was dropped. In this case, it would be 20 meters.
So, the ball bounces to a height of 20 meters.
To find out how high the ball bounces, we can use the concept of conservation of energy.
When the ball is dropped from a height of 20m, it initially has potential energy due to its height above the ground. As it falls, this potential energy is converted into kinetic energy. When it hits the ground, all of its potential energy is converted into kinetic energy.
Upon hitting the ground, the ball rebounds with a speed that is one half of the speed at which it hits the ground. This means that the kinetic energy after the bounce is only a quarter (1/2)^2 of the kinetic energy before the bounce.
Since kinetic energy is directly proportional to the square of the velocity, the ratio of the kinetic energy after the bounce to the initial kinetic energy is (1/2)^2 = 1/4.
Now, since the ball bounces, it can achieve a maximum height before it starts falling again. At this maximum height, the ball has no kinetic energy (because it's momentarily at rest) and all of its energy is in the form of potential energy.
Using the conservation of energy, we can equate the potential energy at the maximum height to the initial kinetic energy before the bounce:
Potential energy at maximum height = Initial kinetic energy
mgh = 1/4 * mgh
In this equation, m represents the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the maximum height.
We can cancel out the mass and 'g' in the equation, and solve for 'h':
h = 1/4 * h
Simplifying further:
3/4 * h = 0
This equation tells us that the maximum height after the bounce is 0. Thus, the ball does not bounce back up after hitting the ground.
Therefore, the height to which the ball bounces is 0 meters.