Question

1. physics when serving in tennis, a player tosses the tennis ball vertically in the air. the height h of the ball after t seconds is given by the quadratic function h(t)= - 5t² + 7t (the height is measured in meters from the point of the toss). a. how high in the air does the ball go? b. assume that the player hits the ball on its way down when its 0.6 m above the point of the toss. for how many seconds is the ball in the air between the toss and the serve?

Explanation:

Step1: Find the vertex of the quadratic function for part a

The quadratic function is $h(t)=-5t^{2}+7t$, where $a = - 5$, $b = 7$, $c = 0$. The $t$-coordinate of the vertex of a quadratic function $y = ax^{2}+bx + c$ is $t=-\frac{b}{2a}$.
$t=-\frac{7}{2\times(-5)}=\frac{7}{10}=0.7$.

Step2: Calculate the maximum height

Substitute $t = 0.7$ into $h(t)$.
$h(0.7)=-5\times(0.7)^{2}+7\times0.7=-5\times0.49 + 4.9=-2.45+4.9 = 2.45$ meters.

Step3: Solve for $t$ when $h(t)=0.6$ for part b

Set $-5t^{2}+7t=0.6$, then $5t^{2}-7t + 0.6=0$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 5$, $b=-7$, $c = 0.6$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-7)^{2}-4\times5\times0.6=49 - 12 = 37$.
Then $t=\frac{7\pm\sqrt{37}}{10}$.
$t_1=\frac{7+\sqrt{37}}{10}\approx\frac{7 + 6.083}{10}=1.3083$ and $t_2=\frac{7-\sqrt{37}}{10}\approx\frac{7 - 6.083}{10}=0.0917$.
The time the ball is in the air between the toss and the serve (on the way - down) is $t_1 - t_2=\frac{7+\sqrt{37}}{10}-\frac{7 - \sqrt{37}}{10}=\frac{\sqrt{37}}{5}\approx1.2166$ seconds.

Answer:

a. $2.45$ meters
b. Approximately $1.22$ seconds