Question

A 70 kg boy is about to dive 5 meters into the water. How much kinetic energy will he have when he is 2 yards above the water?

Answer 1

initial height of the boy when he jump or dive is 5 meter

`h_1 = 5 m`

now his final position is 2 yards above the surface

`h_2 = 2 yards`

as we know that

`1 yard = 0.9144 m`

`2 yards = 1.83 m`

now by energy conservation we can say

change in potential energy = gain in kinetic energy

`mg(h_1 - h_2) = (1)/(2) mv^2`

divide both sides by mass "m"

`g*(5 - 1.83) = (1)/(2)*v^2`

`v^2 = 2*9.8*(5 - 1.83)`

Now kinetic energy will be given as

`KE = (1)/(2) mv^2`

`KE = (1)/(2)*70 * 2*9.8*( 5 - 1.83)`

`KE = 2175 J`

so his kinetic energy will be 2175 J