First, let us consider an object launchedvertically upward with an initial speed v.Neglect air resistance.1. As the projectile goes upward, what energy changes takeplace?Both kineticand potential energy decrease.Both kineticand potential energy increase.Kineticenergy decreases; potential energy increases.Kineticenergy increases; potential energy decreases. 2. Using conservation of energy,find the maximum height h_max to which the object will rise.Express your answer in terms ofvand the magnitude of the acceleration of gravity g.3. At what height h above the ground does the projectile have a speed of0.5v?Express your answer in terms ofvand the magnitude of the acceleration of gravity g.4. What is the speed u of the object at the height of (1/2)h_{\rm max} ?Express your answer in terms ofvand g.Use three significant figures in the numeric coefficient.

1. When a body is thrown vertically all of its mechanical energy is kinetic energy, which is associated with body velocity, as its velocity increases, therefore its kinetic energy decreases.

the potential energy is associated with the height of the body, this is zero on the surface and as the height increases it increases.

Mechanical energy is the sum of these two energies and remains constant throughout the trajectory, because the force of friction is zero.

The correct statement is: The kinetic energy decreases; Increase potential energy.

-2.

Initial Em₀ = K = ½ m v²

Final Emf = U = m g y

Em₀ = Emf

½ m v² = m g y

y = ½ v² / g

3)

y = ½ 0.5² /9.8

y = 0.01276 m

4) if y = [tex]y_{max}[/tex]/ 2

v = √ 2yg

v = √ 2 [tex]y_{max}[/tex]/2 g

v = √ [tex]y_{max}[/tex] g

v = √ 0.01276 9.8

v = 1.11 m / s