201907251758 Solution to 1980-CE-PHY-II-5

The kinetic energy $E_\mathrm{k}$ of an object of mass $m$ and speed $v$ is given by the formula

$E_\mathrm{k}=\displaystyle{\frac{1}{2}}mv^2$ .

In the situation that the object is thrown upwards with initial speed $u$ , and subjected only to gravity $\mathbf{g}$ , it can be expected that after some time of flight $T$ , the object will return to its initial position, its downward speed in which is equal to the initial upward speed $u$ .

Define a piecewise scalar function $v(t)$ of time $t$ :

$v(t) = \begin{cases} u-gt & \quad \textrm{when }0\leq t\leq \displaystyle{\frac{T}{2}} \\ -u+gt & \quad \textrm{when } \displaystyle{\frac{T}{2}}\leq t\leq T \end{cases}$

or simply

$v:[0,T]\subset \mathbb{R} \rightarrow [0,u]\subset \mathbb{R}$ given by $t\mapsto \big|u-g(T-t)\big|$ .

Then

$\begin{aligned} v^2 & =(u-gt)^2\quad \big( =(-u+gt)^2\big) \\ & = u^2-2ugt+g^2t^2 \quad \big(\forall\, t\in [0,T] \big) \end{aligned}$ .

Thus the kinetic energy $E_\mathrm{k}(t)$ is

$\begin{aligned} E_\mathrm{k}(t) & =\displaystyle{\frac{1}{2}}m(u^2-2ugt+g^2t^2) \\ & = \bigg( \displaystyle{\frac{1}{2}}mg^2 \bigg) t^2 + ( -mug ) t + \bigg( \displaystyle{\frac{1}{2}}mu^2 \bigg) \end{aligned}$

where $m$ , $u$ , $g$ are constants.

The kinetic energy $E_\mathrm{k}(t)$ is set to zero at time $t'$ :

$\begin{aligned} t' & = \displaystyle{\frac{-(-mug)\pm\sqrt{\big( -mug\big)^2-4\big(\frac{1}{2}mg^2\big)\big(\frac{1}{2}mu^2\big)}}{2\big(\frac{1}{2}mg^2\big)}} \\ & = \displaystyle{\frac{mug\pm\sqrt{m^2u^2g^2-m^2u^2g^2}}{mg^2}} \\ & =\displaystyle{\frac{u}{g}} \end{aligned}$