Respuesta :
The difference in height between both players, Arabella and Boris, is given by [tex]D(t) = h_A(t) - h_B (t)[/tex]
Further explanation
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:
[tex]h_A (t) = V \cdot t - \frac{g \cdot t^2}{2}[/tex]
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 [tex]0\leq t\leq \frac{2 \cdot V}{g}[/tex]. 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 [tex]t_r[/tex]:
[tex]h_B (t) = V \cdot (t-t_r) - \frac{g \cdot (t-t_r)^2}{2}[/tex]
Which is valid over the interval [tex]t_r \leq t\leq t_r + \frac{2 \cdot V}{g}[/tex], 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:
[tex]D(t) = h_A(t) - h_B (t) = V \cdot t - \frac{g \cdot t^2}{2} [/tex] for all [tex]0 \leq t \leq t_r[/tex]
[tex]D(t) = h_A(t) - h_B (t) = V \cdot t - \frac{g \cdot t^2}{2} - (V \cdot (t-t_r) - \frac{g \cdot (t-t_r)^2}{2}) = V \cdot t_r - g \cdot t_r \cdot t + \frac{g \cdot t_r^2}{2} [/tex] for all [tex]t_r \leq t \leq \frac{2 \cdot V}{g}[/tex]
[tex]D(t) = h_A(t) - h_B (t) = - V \cdot (t-t_r) + \frac{g \cdot (t-t_r)^2}{2} [/tex] for all [tex]\frac{2 \cdot V}{g} \leq t \leq t_r + \frac{2 \cdot V}{g}[/tex]
Learn more
Here is a list of well explained free falling problems, similar to yours:
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Keywords
Free fall, parabolic motion, kinematics
Vertical displacement when [tex]t=t_r[/tex] is [tex]\b{\boxed{\sqrt{2gH}\left(t\right)-\dfrac{1}{2}g{t^2}}}[/tex] and vertical displacement when [tex]{t_r}<t[/tex] is [tex]\b{\fbox{\begin\{t({\sqrt{2gH}+\dfrac{g(t_r-2t)}{2})}\end{minispace}}}[/tex].
Further Explanation:
Arabella and Boris are of same height and jumps with the same initial velocity. Therefore, both can jump to height [tex]H[/tex].
The energy of both the players Arabella and Boris at ground level and height [tex]H[/tex] is constant.
Concept:
Applying conservation of energy at ground and height [tex]H[/tex]:
[tex]mgH=\dfrac{1}{2}m{v_0}^2[/tex]
Rearrange the above equation for initial velocity [tex]v_0[/tex]:
[tex]v_0=\sqrt{2gH}[/tex]
To calculate the height of both players, we can apply the equation of law of motion.
For a given initial velocity [tex]v_0[/tex], acceleration [tex]g[/tex], the height of the Arabella in time [tex]t[/tex]:
[tex]\boxed{{h_a}\left(t\right) ={h_0}+{v_0}t-\dfrac{1}{2}g{t^2}}[/tex]
The reaction time for Boris is [tex]t_r[/tex]. Therefore, to reach the height time taken by her is [tex](t-t_r)[/tex] .
For a given initial velocity [tex]v_0[/tex], acceleration [tex]g[/tex], the height of the Boris in time [tex](t-t_r)[/tex]:
[tex]\boxed{{h_b}(t)={h_0}+{v_0}(t-t_r)+\dfrac{1}{2}g(t-t_r)^2}[/tex]
It is given that Boris jumps before Arabella reaches the maximum height. Therefore, reaction time for Boris cannot be greater than total time.
Case I:
For [tex]t=t_r[/tex], we get [tex]{h_b}(t)={h_0}[/tex] .
Therefore, the vertical displacement is:
[tex]\begin{aligned}D(t)&={h_a}(t)-{h_b}(t)\\&={h_0}+{v_0}t-\dfrac{1}{2}g{t^2}-{h_0}\\&={v_o}t-\dfrac{1}{2}g{t^2}\\&=\sqrt{2gH}(t)-\dfrac{1}{2}g{t^2}\end{gathered}[/tex]
Case II:
For [tex]{t_r}<t[/tex], the vertical displacement is:
[tex]\begin{aligned}D(t)&={h_a}(t)-{h_b}(t)\\&={t_r}({\sqrt{2gH}+\dfrac{g(t_r-2t}{2})\end{aligned}[/tex]
Thus, vertical displacement when [tex]t=t_r[/tex] is [tex]\b{\boxed{\sqrt{2gH}\left(t\right)-\dfrac{1}{2}g{t^2}}}[/tex] and vertical displacement when [tex]{t_r}<t[/tex] is [tex]\b{\fbox{\begin\{t({\sqrt{2gH}+\dfrac{g(t_r-2t)}{2})}\end{minispace}}}[/tex].
Learn more:
1. Average translational kinetic energy: https://brainly.com/question/9078768
2. Stress on a body under tension: https://brainly.com/question/12985068
3. Water is a compound: https://brainly.com/question/4636675
Answer Details:
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords:
Two, basketball, player, same, height, initial, velocity, vertical, distance, Arabella, Boris, high, above, D(t)=Ha(t) – Hb(t), function, reaction, ground, reaches and time.
