Question

Will mark as brainliest:A ball is thrown from an initial height of 3 meters with an initial upward velocity of 25 m/s. The ball's helght h (In meters) after t seconds is given by the following. h=3+25t-5t Find all values of t for which the ball's height is 13 meters, Round your answer(s) to the nearest hundredth. (If there is more than one answer, use the "or" button.)

Answer 1

t = 4.56 sec or 0.44 sec

Explanation:

The equation:

h = 3 + 25t - 5t²

If height = 13 meters, we need to represent h with 13 to get time in seconds

13 = 3 + 25t - 5t²

13 -3 = 25t - 5t²

10 = 2t - 25t²

5t²- 25t + 10 = 0

using quadratic formula:

$\begin{gathered} t\text{ = }\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ a\text{ = 5, b = -25, c = 10} \end{gathered}$$\begin{gathered} t\text{ = }\frac{-(-25)\pm\sqrt[]{(-25)^2-4(5)(10)}}{2(5)} \\ t\text{ = }\frac{25\pm\sqrt[]{625-200}}{10} \end{gathered}$$\begin{gathered} t\text{ = }\frac{25\pm\sqrt[]{425}}{10}\text{ =}\frac{25\pm\sqrt[]{25\times17}}{10} \\ t\text{ = }\frac{25\pm5\sqrt[]{17}}{10}=\frac{5\pm\sqrt[]{17}}{2} \end{gathered}$

In decimal:

$\begin{gathered} t\text{ = }\frac{5+\sqrt[]{17}}{2}or\frac{5-\sqrt[]{17}}{2} \\ t\text{ = }\frac{9.12}{2}\text{ sec or }\frac{0.88}{2} \\ t\text{ = 4}.56sec\text{ or or 0.44sec} \end{gathered}$