A ball is thrown from an initial height of 1 meter with an initial upward velocity of 15 m/s. The ball's height h (in meters) after seconds is given by the following.

h = 1+15t-5t^2

Find all values of t for which the ball's height is 9 meters.

Round your answer (s) to the nearest hundredth.

(If there is more than one answer, use the "or" button.)

t = __ seconds

0.69s

2.31s

Step-by-step explanation:

Given in the question an equation

h = 1 + 15t - 5t²

The problem asks for a time when the ball's height will be 9 m. To find these times, I plug 9 in for h and solve for t:

8 = 1 + 15t - 5t²

-5t² + 15t + 1 - 9 = 0

-5t² + 15t - 8 = 0

Using the quadratic equation:

t = - 15 ± √15²-4 (-5) (-8) / 2 (-5)

t = - 15±√65 / - 10

t = - 15 + √65 / - 10 or t = - 15 - √65 / - 10

t = 0.69 seconds or t = 2.31 seconds

So, the ball is at a height of 9m twice: once on the way up after 0.69 seconds and once on the way back down after 2.31 seconds.