Answer: See below
Step-by-step explanation:
[tex]$Given that,Height of ball hit by Phil,$\begin{aligned}f(t) &=(-16 t-2)(t-2) \\&=-16 t^{2}-2 t+16 \times 2 t+4 \\&=-16 t^{2}+30 t+4\end{aligned}[/tex]
[tex]$Part 1 \\ From the graph $B$ all hit by the John hits the court at $t=1 \mathrm{sec}$After hitting the ground position will be zero.Then.$\begin{aligned}0 &=-16 t^{2}+30 t+4 \\& \Rightarrow t=2 \sec \text { or } \mathrm{t}=-\frac{1}{8}\end{aligned}$Time cannot be negative, so the accepeted value is $2 \ \mathrm{sec}$.That is after $2 \sec$ the ball hit by Phill will reach the ground \\Therefore, the ball hit by John will hit the court first at $t=1 \sec$[/tex]
[tex]$Part 2 \\Maximum height reached by the Phill's ball,$f_{\max }=-16 \times\left(\frac{15}{16}\right)^{2}+30 \times \frac{15}{16}+4=18.06 \mathrm{ft}$From the graph, the maximum height reached by John's ball is $6.2 \mathrm{ft}$ $\underline{\text { So }}$ Phil's tennis ball reach a higher maximum than John's tennis ball[/tex]
[tex]$Part 3 \\The maximum height reached by the Phill's ball,$f_{\max }=-16 \times\left(\frac{15}{16}\right)^{2}+30 \times \frac{15}{16}+4=18.06 \mathrm{ft}=18 \mathrm{ft}$[/tex]
