Question

A ball is thrown from a height of 217 feet with an initial downward velocity of 17 ft/s. The ball's height h (infeet) after t seconds is given by the following.h=217-177-161How 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.)secondsorХ5?ground

Answer 1

The equation representing the ball's movement is given as follows;

`h=217-17t-16t^2`

To determine the value of t (that is, time), we shall now make the equation become;

`217-17t-16t^2=0`

This is because when the height is zero, then the ball has actually touched the ground. So making h equal to zero, we now have a quadratic equation and we shall solve it using the quadratic equation formula as follows;

`t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

The variables are,

`\begin{gathered} -16t^2-17t+217=0 \\ a=-16,b=-17,c=217 \end{gathered}`

We now have;

`\begin{gathered} t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ t=\frac{-(-17)\pm\sqrt[]{(-17)^2-4\lbrack-16\rbrack\lbrack217\rbrack}}{2(-16)} \\ t=\frac{17\pm\sqrt[]{289+13888}}{-32} \\ t=\frac{17\pm\sqrt[]{14177}}{-32} \\ t=(17\pm119.0672)/(-32) \\ t=(17+119.0672)/(-32),t=(17-119.0672)/(-32) \\ t=-4.2521,t=3.1896 \end{gathered}`

The quadratic equation now has two solutions.

Note however, that the question requires us to determine the time taken and we know this cannot be a negative value.

Therefore, we shall take t = 3.1896, and rounded to the nearest hundredth this becomes;

`t\approx3.19\text{ (rounded to the nearest hundredth)}`