A diver weighing 750 N dive from a board 10 m above the surface of the pool pf water. Use the conservation of mechanical energy to find the speed at a point 5 m above the water surface, neglecting air friction. (Ans: 9.9 m s⁻¹)



Given:

Weight of diver = W = 750 N

Height of board = h₁ = 10 m

Height of diver above the surface of water = h₂ = 5 m



To Find:

Speed of the diver =  v = ?   


Solution:


By using Law of conservation of energy

Gain in K.E = Loss in P.E

1212 m v² =  m g h

or

v = √2gh  

Where h =  h₂ - h₁ So,

v = √2g(h₂-h₁)  

by putting the corresponding values

v = √2ｘ9.8ms⁻²(10m-5m)

v = √98m²s⁻²


v = 9.9 m s⁻¹ -------------Ans.

Hence speed of diver is at 5 m above the surface of water is 9.9 m s⁻¹


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