while serving the ball at a beach volleyball ...
while serving the ball at a beach volleyball game, a player tosses the ball up from a height of 4 feet at a velocity of 18 feet per second. after it starts to fall back down, the player jumps up and hits the ball at a height of 7 feet. to the nearest tenth of a second, how long is the ball in the air before the player hits it? hint: use the formula h = -16t² + vt + s.
Answer
# Explanation:
## Step1: Identify the values of h, v, s
Given $h = 7$, $v=18$, $s = 4$. Substitute into $h=-16t^{2}+vt + s$. We get $7=-16t^{2}+18t + 4$.
## Step2: Rearrange the equation
Rearrange $7=-16t^{2}+18t + 4$ to the standard quadratic - form $ax^{2}+bx + c = 0$. So, $16t^{2}-18t + 3=0$.
## Step3: Use the quadratic formula
The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 16$, $b=-18$, $c = 3$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-18)^{2}-4\times16\times3=324 - 192 = 132$.
## Step4: Calculate the values of t
$t=\frac{18\pm\sqrt{132}}{32}=\frac{18\pm2\sqrt{33}}{32}=\frac{9\pm\sqrt{33}}{16}$.
$t_1=\frac{9+\sqrt{33}}{16}\approx\frac{9 + 5.745}{16}=\frac{14.745}{16}\approx0.9$
$t_2=\frac{9-\sqrt{33}}{16}\approx\frac{9 - 5.745}{16}=\frac{3.255}{16}\approx0.2$. We take the larger value since the ball goes up first and then comes down to be hit.
# Answer:
$0.9$