Physics-Energy/ conservation of energy

Question

Question: At a water park, Hugo climbs a 15.00m springboard, in 22.00s, to execute his famous THUNDER CANNON BALL. Hugo jumps up at 6.000m/s [up] off the end of the diving board and propels his mass, of 85.00kg towards the water.
calculate the...
1.weight of Hugo,
2.work done getting to the 15.00m board,
3.potential energy gained at boards edge,
4.power generated in climbing the boards edge,
5.kinetic energy in jumping upwards at the boards edge,
6.max height attained above the board,
7.max height above the water,
8.total energy at max height and
9.speed striking the water.

Answer 1

1. There are 2.2 kg in a pound at the surface of the earth, so his weight is 187 pounds.

 

2. Work is force times distance, if the force is constant.  The force he exerts in climbing is mg, where m = 85 kg, and g = 9.8 kg-m/s², so mg = 833 N.  The distance he climbs is 15 m, so the amount of work he does is 833*15 =  12495 J.

 

3. The potential energy from gravity at a height of h is mgh, where m is his mass and again h is 9.8 kg-m/s².  So in climbing to a height of 15 m, he has added 85 * 9.8 * 15 = 12495 J.

 

4. Since he climbed in 22 seconds, his power was 12495 / 22 = 567.95 watts

 

5. The kinetic energy of his jump is 1/2 mv² = 1/2 * 85 * 6² = 1530 J

 

6. 1530 J of kinetic energy is completely converted to potential energy at the top of his jump.  So 1530 = mgh = 85 * 9.8 * h = 833 h.  So h = 1530 / 833 = 1.837 meters above the board.

 

7.  Since the board is 15 m, his maximum height above the water is 16.837 m.

 

8. The total energy at the maximum height is the total potential energy added by the climb (12495 J) plus the potential energy added by his jump (1530 J).  So the total potential energy at the maximum height is 14025.

 

9. The potential energy relative to the height of the water is all converted to kinetic energy when he splashes in.  So 1/2 mv² = 14025.  Solve this for v...

 

1/2 mv² = 14025

1/2 * 85 v² = 14025

v² =330

v = 18.166 m/s