Question

a football is thrown straight up from 4 meters above the ground with an initial velocity of 15 m/s. assuming that gravity pulls it down, its position can be modeled by a quadratic equation with a coefficient of - 5 m/s² for the t² term. approximately when does it hit the ground? after - 0.25 seconds after 3.25 seconds after 15.25 seconds after 1.5 seconds

Explanation:

Step1: Set up the height - time equation

The general form of the height - time equation for vertical motion is $h(t)=at^{2}+vt + h_0$, where $a$ is the acceleration due to gravity, $v$ is the initial velocity, and $h_0$ is the initial height. Given $a=- 5\ m/s^{2}$, $v = 15\ m/s$, and $h_0 = 4\ m$, the equation is $h(t)=-5t^{2}+15t + 4$. When the football hits the ground, $h(t)=0$, so we need to solve the quadratic equation $-5t^{2}+15t + 4=0$.

Step2: Use the quadratic formula

The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For the equation $-5t^{2}+15t + 4=0$, we have $a=-5$, $b = 15$, and $c = 4$. First, calculate the discriminant $\Delta=b^{2}-4ac=(15)^{2}-4\times(-5)\times4=225 + 80=305$.

Step3: Calculate the values of t

Then, $t=\frac{-15\pm\sqrt{305}}{2\times(-5)}=\frac{-15\pm\sqrt{305}}{-10}=\frac{15\pm\sqrt{305}}{10}$. We have two solutions for $t$: $t_1=\frac{15+\sqrt{305}}{10}\approx\frac{15 + 17.46}{10}=\frac{32.46}{10}=3.246\approx3.25$ and $t_2=\frac{15 - \sqrt{305}}{10}\approx\frac{15-17.46}{10}=\frac{-2.46}{10}=-0.246$. Since time cannot be negative in this context, we discard the negative solution.

Answer:

B. after 3.25 seconds