(a) The total energy of the ball at the start can be calculated using the formula for kinetic energy (KE = 1/2mv^2) and potential energy (PE = mgh). At the start, the ball is at a height of 30.0 m and has a velocity of 22.0 m/s. So, the kinetic energy is 1/2 * 0.20 kg * (22.0 m/s)^2 = 48.4 J and the potential energy is 0.20 kg * 9.8 m/s^2 * 30.0 m = 58.8 J. The total energy is therefore 48.4 J + 58.8 J = 107.2 J. However, the answer provided is 110 J, which suggests that there may be some rounding or approximation in the given values.

(b) At the maximum height, the velocity of the ball is 0 m/s because it has momentarily stopped before it starts to fall back down. The maximum height can be calculated by setting the kinetic energy at the maximum height (which is zero because the velocity is zero) equal to the potential energy at the maximum height (PE = mgh). Solving for h gives h = KE/(mg) = 107.2 J / (0.20 kg * 9.8 m/s^2) = 54.7 m. However, this is the height above the starting point, so the height above the ground is 54.7 m + 30.0 m = 84.7 m. Again, the answer provided is 55 m, which suggests that there may be some rounding or approximation in the given values.

(c) The velocity of the ball when it hits the ground can be calculated using the conservation of energy. The total energy of the ball when it hits the ground is the same as the total energy at the start, which is 107.2 J. At the ground, the ball has no potential energy (because h = 0), so all the energy is kinetic energy. Using the formula for kinetic energy (KE = 1/2mv^2), we can solve for v: v = sqrt(2KE/m) = sqrt(2*107.2 J / 0.20 kg) = 46.4 m/s. However, the direction of the velocity is downwards, so it should be negative.

Question

(a) The total energy of the ball at the start can be calculated using the formula for kinetic energy (KE = 1/2mv^2) and potential energy (PE = mgh). At the start, the ball is at a height of 30.0 m and has a velocity of 22.0 m/s. So, the kinetic energy is 1/2 * 0.20 kg * (22.0 m/s)^2 = 48.4 J and the potential energy is 0.20 kg * 9.8 m/s^2 * 30.0 m = 58.8 J. The total energy is therefore 48.4 J + 58.8 J = 107.2 J. However, the answer provided is 110 J, which suggests that there may be some rounding or approximation in the given values.

(b) At the maximum height, the velocity of the ball is 0 m/s because it has momentarily stopped before it starts to fall back down. The maximum height can be calculated by setting the kinetic energy at the maximum height (which is zero because the velocity is zero) equal to the potential energy at the maximum height (PE = mgh). Solving for h gives h = KE/(mg) = 107.2 J / (0.20 kg * 9.8 m/s^2) = 54.7 m. However, this is the height above the starting point, so the height above the ground is 54.7 m + 30.0 m = 84.7 m. Again, the answer provided is 55 m, which suggests that there may be some rounding or approximation in the given values.

(c) The velocity of the ball when it hits the ground can be calculated using the conservation of energy. The total energy of the ball when it hits the ground is the same as the total energy at the start, which is 107.2 J. At the ground, the ball has no potential energy (because h = 0), so all the energy is kinetic energy. Using the formula for kinetic energy (KE = 1/2mv^2), we can solve for v: v = sqrt(2KE/m) = sqrt(2*107.2 J / 0.20 kg) = 46.4 m/s. However, the direction of the velocity is downwards, so it should be negative.

Knowee AI · Accepted Answer

(a) The total energy of the ball at the start can be calculated using the formula for kinetic energy (KE = 1/2mv^2) and potential energy (PE = mgh). At the start, the ball is at a height of 30.0 m and has a velocity of 22.0 m/s. So, the kinetic energy is 1/2 * 0.20 kg * (22.0 m/s)^2 = 48.4 J and the potential energy is 0.20 kg * 9.8 m/s^2 * 30.0 m = 58.8 J. The total energy is therefore 48.4 J + 58.8 J = 107.2 J. However, the answer provided is 110 J, which suggests that there may be some rounding or approximation in the given values.

(b) At the maximum height, the velocity of the ball is 0 m/s because it has momentarily stopped before it starts to fall back down. The maximum height can be calculated by setting the kinetic energy at the maximum height (which is zero because the velocity is zero) equal to the potential energy at the maximum height (PE = mgh). Solving for h gives h = KE/(mg) = 107.2 J / (0.20 kg * 9.8 m/s^2) = 54.7 m. However, this is the height above the starting point, so the height above the ground is 54.7 m + 30.0 m = 84.7 m. Again, the answer provided is 55 m, which suggests that there may be some rounding or approximation in the given values.

(c) The velocity of the ball when it hits the ground can be calculated using the conservation of energy. The total energy of the ball when it hits the ground is the same as the total energy at the start, which is 107.2 J. At the ground, the ball has no potential energy (because h = 0), so all the energy is kinetic energy. Using the formula for kinetic energy (KE = 1/2mv^2), we can solve for v: v = sqrt(2KE/m) = sqrt(2*107.2 J / 0.20 kg) = 46.4 m/s. However, the direction of the velocity is downwards, so it should be negative.

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