a baseball is hit straight upwards with a velocity of 96 feet per second. the height of the baseball, in…

Explanation:

Step1: Find the time of maximum height

The height - function is a quadratic function $h(t)=-16t^{2}+96t + 3$, where $a=-16$, $b = 96$, $c = 3$. For a quadratic function $y = ax^{2}+bx + c$, the $t$ - value of the vertex (where maximum or minimum occurs) is given by $t=-\frac{b}{2a}$. $t=-\frac{96}{2\times(-16)}=\frac{-96}{-32}=3$ seconds.

Step2: Find the maximum height

Substitute $t = 3$ into the height - function $h(t)$. $h(3)=-16\times3^{2}+96\times3 + 3=-16\times9+288 + 3=-144+288 + 3=147$ feet.

Step3: Find the time when the baseball hits the ground

Set $h(t)=0$, so we have the quadratic equation $-16t^{2}+96t + 3=0$. Use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=-16$, $b = 96$, $c = 3$. First, calculate the discriminant $\Delta=b^{2}-4ac=(96)^{2}-4\times(-16)\times3=9216 + 192=9408$. Then $t=\frac{-96\pm\sqrt{9408}}{2\times(-16)}=\frac{-96\pm\sqrt{16\times588}}{-32}=\frac{-96\pm4\sqrt{588}}{-32}=\frac{-96\pm4\times14\sqrt{3}}{-32}=\frac{-96\pm56\sqrt{3}}{-32}$. We take the positive root since time cannot be negative. $t=\frac{-96 + 56\sqrt{3}}{-32}=\frac{96-56\sqrt{3}}{32}=\frac{6 - 3.5\sqrt{3}}{2}\approx\frac{6-3.5\times1.732}{2}=\frac{6 - 6.062}{2}\approx6.03$ seconds.

Answer:

Maximum height: 147 feet Time of maximum height: 3 seconds Time when it hits the ground: approximately 6.03 seconds