Question

a basketball player throws a basketball with an initial vertical velocity of 48 feet per second from a height of 5 feet. use the vertical motion model, $h = - 16t^{2}+vt + s$ where $v$ is the initial velocity in feet per second and $s$ is the height in feet, to calculate how long the basketball will be in the air for. round your answer to the nearest tenth. time in the air: ______ seconds enter the answer

Explanation:

Step1: Identify the values of v and s

Given $v = 48$, $s=5$, and when the ball hits the ground $h = 0$. So the equation becomes $0=-16t^{2}+48t + 5$.

Step2: Use the quadratic formula

The quadratic formula for $ax^{2}+bx + c=0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here $a=-16$, $b = 48$, $c = 5$. First, calculate the discriminant $\Delta=b^{2}-4ac=(48)^{2}-4\times(-16)\times5=2304 + 320=2624$.

Step3: Calculate the values of t

$t=\frac{-48\pm\sqrt{2624}}{-32}=\frac{-48\pm51.225}{-32}$. We get two solutions for t: $t_1=\frac{-48 + 51.225}{-32}=\frac{3.225}{-32}\approx - 0.1$ (rejected since time cannot be negative) and $t_2=\frac{-48-51.225}{-32}=\frac{-99.225}{-32}\approx3.1$.

Answer:

$3.1$