You throw a 238 g baseball straight up from a height of 1.73 m and it reaches a height of 13.2 m. Using conservation of energy, what minimum initial speed is needed to accomplish this?
As the problem states, this is a conservation of energy problem, so
E(initial) = E(final)
Let's pick our coordinate system to be point at which the ball initially rests. This makes the initial potential energy 0. When the ball is at its highest point, the velocity is temporarily 0, so at its highest point, the kinetic energy is 0. These simplify our equation to
1/2mv(i)^2 = mgh(f)
Plugging in the given values:
(1/2)(.238)(v(i)^2) = (.238)(9.8)(1.73)
v(i) = 5.82 m/s