Question

At the moment when a shot-putter releases a 5.00kg shot, the shot is 3.00 m above the ground and traveling at 15.0m/s. It reaches a maximum height of 14.5 m above the ground and then falls to the ground. i) What was the kinetic energy of the shot as it left the hand? ii)What was the total energy of the shot as it left the hand? iii) What was the total energy of the shot at its maximum height? iv) What was the potential energy of the shot at its maximum height? v) What was the kinetic energy of the shot at its maximum height? vi) What was the kinetic energy of the shot just as it struck the ground? Guys, it would be really cool if I got the solutions instead of just answers.
explaination also needed ​

Answer 1

Answer:

1. 562.5 J

2. 562.5 J

Explanation:

1. Kinetic energy = 1/2 m$v^{2}$ = $\frac{1}{2} * 5 * 15^{2}$ = 562.5

2. According to law of conservation of energy, the total energy is conserved. (Kinetic Energy)

3. According to law of conservation of energy, the total energy is conserved. (PotentialEnergy)

4. Answer 3 and 2

5. Answer 3 and 2

6. Answer 1