Two basketball players are essentially equal in all respects. In particular, by ju

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This problem is a 1-dimensional free fall type of problem, it's one dimensional since both players only move in 1 dimension (meaning vertically), and it is a free fall problem since, when both players leave the ground, the only force which is apllied over them is the force excerted by gravity.

To solve our problem, let's suppose that both players jump from a point called the origin, meaning a point where we consider that their vertical displacement is 0 (in this case, the floor would be considered as the origin). Besides that, let's suppose that we start counting the time at the moment when Arabella makes her jump, and let's say that both players jump with velocity V. Therefor from the equations from kinematics about free falling objects we can write an expression for the height of Arabella over time as:

$h_A(t)=V\cdot t-\frac{g\cdot t^2}{2}$

Which is an equation valid since the time Arabella jumps until she gets back on the ground, in math terms this would be written as that the equation is valid over the interval $0\le t\le \frac{2V}{g}$ At all other times the height of Arabella would be zero (since she is on the ground).

We can write the same for Boris, though his equation is shifted in time by his reaction time $t_r$:

$h_B(t)=V\cdot (t-t_r)-\frac{g\cdot (t-t_r{)}^2}{2}$

Which is valid over the interval $t_r\le t\le t_r+\frac{2V}{g}$ , and zero everywhere else (since Boris is touching the ground at all other moments). Putting all the pieces together, we can get a piecewise function that expresses the difference in height between both players:

$D(t)=h_A(t)-h_B(t)=V\cdot t-\frac{g\cdot t^2}{2}$ for all $0\le t\le t_r$

$D(t)=h_A(t)-h_B(t)=V\cdot t-\frac{g\cdot t^2}{2}-(V\cdot (t-t_r)-\frac{g(t-t_r{)}^2}{2})=V\cdot t_r-g\cdot t_r\cdot t+\frac{g{t}_r^2}{2}$ for all $t_r\le t\le \frac{2V}{g}$

$D(t)=h_A(t)-h_B(t)=-V\cdot (t-t_r)+\frac{g\cdot (t-t_r{)}^2}{2}$ for all $\frac{2V}{g}\le t\le t_r+\frac{2V}{g}$