Question

you throw a volleyball upward with an initial vertical velocity of 3 feet per second from a building that is 18 feet tall. use the vertical motion model, $h = - 16t^{2}+vt + s$ where $v$ is the initial velocity in feet per second and $s$ is the initial height in feet, to calculate how long the volleyball will be in the air for. round your answer to the nearest tenth. time in the air: ______ seconds

Explanation:

Step1: Identify the values of v and s

Given $v = 30$ (assuming the initial velocity is 30 feet - per - second as the number is cut off in the image), $s=18$, and when the volleyball hits the ground $h = 0$. Substitute into the equation $h=-16t^{2}+vt + s$, we get $0=-16t^{2}+30t + 18$.

Step2: Rewrite the equation in standard quadratic form

Multiply through by - 1 to get $16t^{2}-30t - 18 = 0$. Divide by 2 to simplify: $8t^{2}-15t - 9=0$.

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, $a = 8$, $b=-15$, $c=-9$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-15)^{2}-4\times8\times(-9)=225 + 288=513$.

Step4: Calculate the values of t

$t=\frac{15\pm\sqrt{513}}{16}$. $\sqrt{513}\approx22.65$. So $t=\frac{15\pm22.65}{16}$. We have two solutions for t: $t_1=\frac{15 + 22.65}{16}=\frac{37.65}{16}\approx2.4$ and $t_2=\frac{15-22.65}{16}=\frac{-7.65}{16}\approx - 0.5$. Since time cannot be negative, we discard the negative solution.

Answer:

$2.4$