a ball is thrown from a height of 79 meters with an initial downward velocity of 6 m/s. the balls height h…

a ball is thrown from a height of 79 meters with an initial downward velocity of 6 m/s. the balls height h (in meters) after t seconds is given by the following. h = 79 - 6t - 5t² 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.)

a ball is thrown from a height of 79 meters with an initial downward velocity of 6 m/s. the balls height h (in meters) after t seconds is given by the following. h = 79 - 6t - 5t² 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.)

Answer

Explanation:

Step1: Set height h to 0

When the ball hits the ground, $h = 0$. So we set up the equation $0=79 - 6t-5t^{2}$.

Step2: Rearrange the equation

We rewrite it in standard quadratic - form $5t^{2}+6t - 79 = 0$.

Step3: Apply the quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 5$, $b = 6$, and $c=-79$. First, calculate the discriminant $\Delta=b^{2}-4ac=(6)^{2}-4\times5\times(-79)=36 + 1580 = 1616$.

Step4: Find the values of t

$t=\frac{-6\pm\sqrt{1616}}{10}=\frac{-6\pm40.2}{10}$. We have two solutions: $t_1=\frac{-6 + 40.2}{10}=\frac{34.2}{10}=3.42$ and $t_2=\frac{-6 - 40.2}{10}=\frac{-46.2}{10}=-4.62$. Since time cannot be negative in this context, we discard the negative solution.

Answer:

$3.42$