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=(-b\pm√(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=(-(-12)\pm√((-12)^2-4(-16)(164)))/(2(-16))`
`t_1=(12+√(10640))/(-32)\approx-3.60`
`t_2=(12-√(10640))/(-32)\approx2.85`

Choose the positive value:

`t\approx2.85`

Hence, the answer is:

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