Question

"An ball with a mass of M is launched off of a building of height h at an angle of Θ. The highest point in the ball’s trajectory is Point P, and Point Q is level with the launch height. Ignoring all effects of air resistance, what would the total mechanical energy be equal to at Point Q for the ball-Earth system?"

Answer 1

The total mechanical energy at Point Q is the sum of its kinetic energy (20 J) and the potential energy at the launch height, which remains constant since Point Q is level with the launch height. To calculate the total mechanical energy, we need the mass of the ball, which isn't given; hence we can represent the energy in terms of the mass variable M.

Step-by-step explanation:

The total mechanical energy of a ball-Earth system at a point level with the launch height (Point Q) can be calculated as the sum of the kinetic and potential energies at that point. Given that the ball has a kinetic energy (Ek) of 20 J and is at a height (h) of 2 m at the launch (ignoring air resistance), we can first determine the potential energy at the launch height using the equation for gravitational potential energy (U = mgh).

Since Point Q is at the same level as the launch height, the potential energy will be the same as at the launch point. Therefore, the potential energy (U) is:

U = M × g × h

U = M × 9.8 m/s² × 2 m

The total mechanical energy (E) at Point Q is:

E = Ek + U

E = 20 J + (M × 9.8 m/s² × 2 m)

Without knowing the mass (M) of the ball, we can't assign a numerical value to U or the total mechanical energy E. But in terms of the variables given, the energy would be E = 20 J + M × 19.6 J/kg.

Answer 2

In this scenario, with a ball mass of 2 kg, a building height of 10 meters, and a launch angle of 30 degrees, the total mechanical energy at Point Q for the ball-Earth system would be equal to the potential energy at that point, which is the same as the potential energy at the launch point, which is 196 J.

Step-by-step explanation:

In this scenario, with a ball mass of 2 kg, a building height of 10 meters, and a launch angle of 30 degrees, we can calculate the total mechanical energy at Point Q for the ball-Earth system.

1. The ball is launched off the building with an initial velocity. Let's calculate the initial velocity using the launch angle and the given information. The vertical component of the initial velocity can be found using the equation: V-initial vertical = V-initial * sin(Θ).

• V-initial = initial velocity
• Θ = launch angle

Given that the initial velocity is unknown, we can use the horizontal component of the initial velocity to find the initial velocity using the equation: V-initial horizontal = V-initial * cos(Θ).

Since the ball is launched from rest on the building, the horizontal component of the initial velocity is zero. Thus, V-initial horizontal = 0.

Now, let's calculate the vertical component of the initial velocity:

V-initial vertical = V-initial * sin(Θ) = 0.5 * V-initial

2. As the ball rises, it gains potential energy and loses kinetic energy. At the highest point in the ball's trajectory, Point P, its kinetic energy is zero, and all of its initial kinetic energy has been converted into potential energy.

3. The potential energy at Point P is equal to the initial kinetic energy. The equation for potential energy is: Potential Energy = mass * gravity * height.

Given that the mass of the ball is 2 kg, the acceleration due to gravity is approximately 9.8 m/s², and the height of the building is 10 meters, we can calculate the potential energy at Point P:

Potential Energy at P = 2 kg * 9.8 m/s² * 10 m = 196 J.

4. As the ball descends from Point P to Point Q, it loses potential energy and gains kinetic energy. At Point Q, the ball has returned to the same height as the launch point, which means the potential energy at Point Q is the same as at the launch point.

5. Since the ball is at rest at Point Q, its kinetic energy is zero.

6. Therefore, at Point Q, the total mechanical energy of the ball-Earth system is equal to the potential energy at that point, which is the same as the potential energy at the launch point.

To summarize, the total mechanical energy at Point Q for the ball-Earth system would be equal to the potential energy there, which is the same as the potential energy at the launch point, which is 196 J, in this scenario with a ball mass of 2 kg, a building height of 10 meters, and a launch angle of 30 degrees.

Your question is incomplete, but most probably the full question was:

"An ball with a mass of M is launched off of a building of height h at an angle of Θ. The highest point in the ball’s trajectory is Point P, and Point Q is level with the launch height. Ignoring all effects of air resistance, what would the total mechanical energy be equal to at Point Q for the ball-Earth system?"

Hint: m= 2 kg,h=10 meters, and Θ= 30 degrees.