Hi there!

You've got a fairly challenging question there, but I can help you out .

Both b***s have potential energy since they are at a balcony at a height of 18.8 meters.

The formula for potential energy is given by

$$E_p = mgh$$

where Ep is the potential energy, m is the mass, g is the acceleration of gravity and h is the height.

Generally g is approximated to be about 9.8 (this differs across the globe).

Therefore $$E_p = m*9.8*18.8 = 184.24*m$$

In addition to that we know that the first ball is thrown downwards with a force of 13.1 m/s

The formula for kinetic energy is given by

$$E_k = \frac{1}{2}mv^2$$

where Ek is the kinetic energy, m is the mass and v is the velocity (speed).

Therefore we know this throw has given an additional energy of.

$$E_k = \frac{1}{2}m(13.1)^2 = 85.805*m$$

So the total energy when the ball is thrown downwards is given by

$$E_t = E_k + E_p = 184.24*m+85.805*m = 270.045*m$$

Now by the first law of thermodynamics we know that energy always remains constant.

If we neglect air resistance, all energy should have been transformed into kinetic energy, a fraction of a milisecond before it hits the ground.

Therefore

$$\begin{array}{lcl}

270.045*m = E_k = \frac{1}{2}mv^2\\

270.045 = \frac{1}{2}v^2 \mbox{ (divide both sides by m)}\\

v^2 = 2*270.045\\

v = \sqrt{2*270.045} \approx 23.2 m/s\\

\end{array}$$

So the answer to the first question is 23.2 m/s

Now for the second ball, let's pretend that the balcony is at height 0. That would mean that when the ball is thrown upwards with a speed of 13.1m/s it will reach a certain height. At this height, all kinetic energy will have transformed into potential energy.

If we want to calculate this height we can therefore again equate

$$E_k = E_p$$

$$\begin{array}{lcl}

\frac{1}{2}mv^2 = mgh\\

\frac{1}{2}v^2 = gh\\

h = \frac{v^2}{2g}\\

h = \frac{13.1^2}{2*9.8}\\

h \approx 8.76\mbox{meters}

\end{array}$$

Now let's get rid of the assumption that the balcony is at a height of 0.

We now know that the ball loses all it's kinetic energy at a height of 8.76+18.8 = 27.56 meters

This means at this height all it's energy has become potential energy.

We can calculate this potential energy with the equation

$$E_p = m*g*h = m*9.8*27.56 = 270.088*m$$

Again we know that when the ball is just a fraction from the ground all potential energy will have become kinetic energy.

Hence we can again equate this to

$$E_k = \frac{1}{2}mv^2$$

Solving this gives

$$\begin{array}{lcl}

\frac{1}{2}{m}v^2 = 270.088*m\\

\frac{1}{2}v^2 = 270.088\\

v^2 = 2*270.088\\

v = \sqrt{2*270.088}\\

v \approx 23.2m/s\\

\end{array}$$

Since we have neglected air resistance here you can see that the law of thermodynamics apply.

Since both b***s have the same total energy when they leave the balcony, both b***s will have the same total energy when they hit the ground. Since in both cases this energy has totally become kinetic energy. the velocity of both b***s is the same when they hit the ground.

I'll have to look up some formula's with time before I'll be able to answer the other questions (since I haven't done this in a while).

Reinout

p.s. I might have made some round-off errors along the way since I've rounded off my answer several times during the calculations. My apologies for that.