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

`\frac {1}{2}` m `\v²` =  m g h

or

v = `\sqrt {2gh}`  

Where h =  h₂ - h₁ So,

v = `\sqrt {2g(h₂ - h₁)}`  

by putting the corresponding values

v = `\sqrt {2ｘ 9.8 m s⁻² (10 m - 5 m)}`

v = `\sqrt {98 m² s⁻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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