The ball is launched with total initial energy Einitial = mgho. When it reaches the bottom of the hoops, it now has a potential ernergy of PE mghplat for m. How much kinetic energy must it have at that point? Omgho Omghplat formmgho mghplat form Omgho mghplat for m

  K = m g (h₀ - h_plat)

Explanation:

Let's use energy conservation to solve this problem, write the energy at two points of interest

Initial. Higher

          E_initial = m g h₀

Final. Lower

          E_end = K + Ep

          E_end = ½ m v² + m g h_plat

Energy is conserved

          E_initial = E_end

          m g h₀ = K + m g h_plat

          K = m g (h₀ - h_plat)

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2