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A 0.20 kg ball is thrown straight up from the edge of a 30.0 m tall building at a velocity of 22.0 m/s. the ball moves up to the maximum height and then falls to the ground at the base of the building. Assuming that the reference level for gravitational potential energy is ground level. What is the total energy?

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Final answer:

The total energy of the ball thrown from the building is the sum of its initial kinetic energy and gravitational potential energy. Using the conservation of energy principle, this value remains constant throughout the ball's motion, assuming no energy loss.

Step-by-step explanation:

For a projectile motion problem where we have a 0.20 kg ball thrown from a 30.0 m tall building with a velocity of 22.0 m/s, to find the total energy at the beginning, we'll consider both kinetic and potential energy.

  • Kinetic Energy (KE) at the start can be calculated using the formula KE = (1/2)mv², where 'm' is mass and 'v' is velocity.
  • Gravitational Potential Energy (GPE) at the start can be calculated as GPE = mgh, where 'm' is mass, 'g' is acceleration due to gravity (9.81 m/s²), and 'h' is height.

The initial kinetic energy would therefore be (1/2)*0.20 kg*(22.0 m/s)². The initial potential energy would be 0.20 kg*9.81 m/s²*30.0 m. The total energy is the sum of both kinetic and potential energy at the start. As long as there is no energy loss due to air resistance or other non-conservative forces, this total energy would remain constant throughout the ball's motion due to the Conservation of Energy.

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