Question

a ball is thrown from a height of 40 meters with an initial downward velocity of 5 m/s. the balls height h (in meters) after t seconds is given by the following. h = 40 - 5t - 5t^2. how long after the ball is thrown does it hit the ground? round your answer(s) to the nearest hundredth. (if there is more than one answer, use the \or\ button.)

Explanation:

Step1: Set height to 0

When the ball hits the ground, $h = 0$. So we set up the equation $0=40 - 5t-5t^{2}$.

Step2: Rearrange the equation

Divide the entire equation by - 5 to simplify: $t^{2}+t - 8=0$.

Step3: Apply quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 1$, $b = 1$, and $c=-8$. So $t=\frac{-1\pm\sqrt{1^{2}-4\times1\times(-8)}}{2\times1}=\frac{-1\pm\sqrt{1 + 32}}{2}=\frac{-1\pm\sqrt{33}}{2}$.

Step4: Calculate the values of t

We have two solutions for $t$: $t_1=\frac{-1+\sqrt{33}}{2}$ and $t_2=\frac{-1 - \sqrt{33}}{2}$. Since time cannot be negative in this context, we discard $t_2$. Calculate $t_1=\frac{-1+\sqrt{33}}{2}\approx\frac{-1 + 5.745}{2}=\frac{4.745}{2}=2.3725\approx2.37$.

Answer:

$t = 2.37$ seconds