Respuesta :
initial height of the boy when he jump or dive is 5 meter
[tex]h_1 = 5 m[/tex]
now his final position is 2 yards above the surface
[tex]h_2 = 2 yards[/tex]
as we know that
[tex]1 yard = 0.9144 m[/tex]
[tex]2 yards = 1.83 m[/tex]
now by energy conservation we can say
change in potential energy = gain in kinetic energy
[tex]mg(h_1 - h_2) = \frac{1}{2} mv^2[/tex]
divide both sides by mass "m"
[tex]g*(5 - 1.83) = \frac{1}{2}*v^2[/tex]
[tex]v^2 = 2*9.8*(5 - 1.83)[/tex]
Now kinetic energy will be given as
[tex]KE = \frac{1}{2} mv^2[/tex]
[tex]KE = \frac{1}{2}*70 * 2*9.8*( 5 - 1.83)[/tex]
[tex]KE = 2175 J[/tex]
so his kinetic energy will be 2175 J
