Question

a ball is thrown from a height of 43 meters with an initial downward velocity of 4 m/s. the balls height h (in meters) after t seconds is given by the following. h = 43 - 4t - 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 $0=43 - 4t-5t^{2}$.

Step2: Rearrange to standard quadratic form

We rewrite the equation as $5t^{2}+4t - 43=0$.

Step3: Apply quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 5$, $b = 4$, and $c=-43$. So $t=\frac{-4\pm\sqrt{4^{2}-4\times5\times(-43)}}{2\times5}=\frac{-4\pm\sqrt{16 + 860}}{10}=\frac{-4\pm\sqrt{876}}{10}=\frac{-4\pm2\sqrt{219}}{10}=\frac{-2\pm\sqrt{219}}{5}$.

Step4: Calculate positive value of t

We have two solutions for $t$, but since time cannot be negative in this context, we consider only the positive solution. $t=\frac{-2+\sqrt{219}}{5}$. $\sqrt{219}\approx14.799$. Then $t=\frac{-2 + 14.799}{5}=\frac{12.799}{5}=2.56$ (rounded to the nearest hundredth).

Answer:

$t = 2.56$ seconds