Question

a baseball is hit with an initial upward velocity of 70 feet per second from a height of 4 feet above the ground. the equation h = - 16t² + 70t + 4 models the height in feet t seconds after it is hit. after the ball gets to its maximum height, it comes down and is caught by another player at a height of 6 feet above the ground. about how long after it was hit does it get caught? 0.03 seconds 4.38 seconds 4.43 seconds 4.35 seconds

Explanation:

Step1: Set up the equation

We are given the height - time equation \(h=-16t^{2}+70t + 4\), and we know that the ball is caught at \(h = 6\). So we set up the equation \(6=-16t^{2}+70t + 4\).

Step2: Rearrange the equation to standard quadratic form

Subtract 6 from both sides to get \(0=-16t^{2}+70t + 4 - 6\), which simplifies to \(0=-16t^{2}+70t - 2\). Divide through by -2 to get \(8t^{2}-35t + 1=0\).

Step3: Use the quadratic formula

The quadratic formula for a quadratic equation \(ax^{2}+bx + c = 0\) is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 8\), \(b=-35\), and \(c = 1\). First, calculate the discriminant \(\Delta=b^{2}-4ac=(-35)^{2}-4\times8\times1=1225 - 32=1193\). Then \(t=\frac{35\pm\sqrt{1193}}{16}\).

Step4: Calculate the two values of t

\(t_1=\frac{35+\sqrt{1193}}{16}\approx\frac{35 + 34.54}{16}=\frac{69.54}{16}\approx4.35\) and \(t_2=\frac{35 - \sqrt{1193}}{16}\approx\frac{35 - 34.54}{16}=\frac{0.46}{16}\approx0.03\). The smaller value \(t\approx0.03\) seconds represents the time when the ball is on its way up, and the larger value \(t\approx4.35\) seconds represents the time when the ball is on its way down. Since we want the time when the ball is caught on its way down, we choose \(t\approx4.35\) seconds.

Answer:

4.35 seconds