Question

ball is thrown from a height of 164 feet with an initial downward velocity of 12ft / s The ball's height (in feet) after seconds is given by the following h = 164 - 12t - 16t ^ 2 How long after the ball is thrown does it hit the ground? Round your answer(s) to the nearest hundredth there is more than one answer use the or" button )

Answer 1

Given that the ball's height (in feet) after seconds is given by the following equation:

$h=164-12t-16t^2$

You can determine that when the ball hits the ground:

$h=0$

Then, you can substitute that value into the equation:

$0=164-12t-16t^2$

In order to find the values of "t", you can follow these steps:

1. Rewrite it in the form:

$ax^2+bx+c=0$

Then:

$-16t^2-12t+164=0$

2. Use the Quadratic Formula:

$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case:

$\begin{gathered} a=-16 \\ b=-12 \\ c=164 \end{gathered}$

Therefore, you can substitute values into the formula and evaluate:

$t=\frac{-(-12)\pm\sqrt{(-12)^2-4(-16)(164)}}{2(-16)}$$t_1=\frac{12+\sqrt{10640}}{-32}\approx-3.60$$t_2=\frac{12-\sqrt{10640}}{-32}\approx2.85$

Choose the positive value:

$t\approx2.85$

Hence, the answer is:

$t\approx2.85\text{ }seconds$