Question

with what initial speed must you throw a ball vertically for it to reach a height of 20m? the ball weighs 15kg. how does your answer change if the ball weighs 30kg

Answer 1

In a closed system, energy is constant. Thus, the sum of changes in potential and kinetic energy is zero: \[\Delta KE+\Delta PE=0\]where\[\Delta KE = \frac{1}{2}m(v_{f}^{2}-v_{i}^{2})\]and \[\Delta PE = mg(h_{f}-h_{i})\]where m is mass, v is velocity, g is acceleration due to gravity, and h is the height.

Initially (the moment you throw the ball upwards), the height of the ball is at its lowest point (h=0). Consequently, the potential energy is zero and the only energy is kinetic. When the ball has reached its maximum height, v=0. Thus, the only energy is potential.

Putting this together, we get \[-\frac{1}{2}mv_{i}^{2}+mgh_{f}=0\] or \[\frac{1}{2}mv_{i}^{2}=mgh_{f}\] Notice that mass appear in both terms and cancels. Solving for initial velocity, we get\[v_{i}=\sqrt{2gh_{f}}\]

Note that this result is independent of mass, so doubling it will have no affect on the initial velocity required.