Question

(e) On another occasion, the boy throws the ball down from a height of 1.1 m, giving an initailkinetic energy of 9.0 JCalculate the speed at which the ball is the ground​

Answer 1

The speed at which the ball hits the ground is approximately 4.2 m/s.

Step-by-step explanation:

To find the speed at which the ball hits the ground, we can use the principle of conservation of energy. The initial kinetic energy (9.0 J) is converted into gravitational potential energy as the ball falls. The gravitational potential energy at the initial height is given by mgh , where m is the mass, g is the acceleration due to gravity, and h is the height. Equating the initial kinetic energy to the gravitational potential energy:

`\[ mgh = (1)/(2)mv^2 \]`

Canceling the mass m from both sides and rearranging the equation, we get:

`\[ v = √(2gh) \]`

Plugging in the values
`(\( g = 9.8 \, \text{m/s}^2, \, h = 1.1 \, \text{m} \))`, we find:

`\[ v = √(2 * 9.8 * 1.1) \approx 4.2 \, \text{m/s} \]`

This is the speed at which the ball hits the ground.

Answer 2

The final speed at which the ball hits the ground, considering the initial kinetic energy and the potential energy converted back into kinetic energy, is approximately
`\(6.57 \text{ m/s}\).`

To find the speed at which the ball hits the ground, we can use the conservation of energy principle. The initial kinetic energy given to the ball will be converted into gravitational potential energy as it rises to its maximum height, and then back into kinetic energy as it falls.

Given:

- Initial kinetic energy,
`\( KE_{\text{initial}} = 9.0 \)`Joules

- Gravitational potential energy at the height, PE = mgh , where:

- m is the mass of the ball (not given, but we can find it from the initial kinetic energy and the initial speed)

- g is the acceleration due to gravity (\( g = 9.81 \text{m/s}^2 \))

- h is the height (\( h = 1.1 \) meters)

The total mechanical energy at any point in the motion will be the sum of its potential and kinetic energies. At the top of its trajectory, all the kinetic energy will have been converted into potential energy. As it falls, this potential energy will convert back into kinetic energy. When the ball hits the ground, all the energy will be kinetic.

Step 1: Calculate the mass of the ball using the initial kinetic energy and the formula for kinetic energy:

`\[ KE_{\text{initial}} = (1)/(2) m v_{\text{initial}}^2 \]`

We don't have the initial speed, so we will first calculate the mass with the height and the potential energy at that height.

Step 2: Since the ball is thrown down from a height of 1.1 m, the potential energy at that height will be equal to the initial kinetic energy (since the ball will momentarily stop at that height if thrown upwards with that kinetic energy). Therefore,

`\[ PE = KE_{\text{initial}} \]`

`\[ mgh = KE_{\text{initial}} \]`

We can solve for m from this equation.

Step 3: Once we have the mass, we can use the conservation of energy to find the speed at which the ball hits the ground. The potential energy at the height will be converted into kinetic energy as the ball falls:

`\[ PE + KE_{\text{initial}} = KE_{\text{final}} \]`

`\[ mgh + KE_{\text{initial}} = (1)/(2) m v_{\text{final}}^2 \]`

Since the initial kinetic energy is the energy with which the ball is thrown downwards, we can add it to the potential energy to get the total energy just before the ball starts falling.

Step 4: Solve for the final speed
`\( v_{\text{final}} \)`using the above equation.

Let's go through the calculations.

The mass of the ball, calculated using the given potential energy at the height of 1.1 meters and the initial kinetic energy of 9.0 Joules, is approximately
`\(0.834 \text{ kg}\).`