Question

2. a 755N diver drops from a board 10.0 m above the water's surface. Find the divers speed 5.00 m above the water's surface. then find the diver's speed just before striking the water.

3. if the diver in item 2 leaves the board with an initial upward speed of 2.00 m/s find the divers speed when striking the water.

i don't know how to start these two at all...i don't need the answer i just need the formulas that i use for it...i don't understand which ones to use.. plz help

Answer 1

Use a conservation of energy approach for both problems. The CHANGE in (1/2)M V^2 equals the change in M g H.

That means the change in V^2 equals 2 g times the change in height above the water, H.

For the second problem, the water is 10 m below but he had a speed of V^2 = 2 m/s at the start.

Vfinal^2 = Vinitial^2 + 2 g h
= 4.0 + 2*9.8*10 = 200
V = 10 m/s

Answer 2

I don’t understand

Answer 3

I don’t understand can you show me