A ball is thrown from a height of 105 feet with an initial downward velocity of 9 ft/s. The ball's height h (in feet) after t seconds is given by the following.

h = 105 - 9t-16t^2

How 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.)

1 =

seconds

X

5

?

ground

t = 2.28 s

Step-by-step explanation:

h = 105 - 9t - 16t ^ 2

0 ft = 105 ft - 9t - 16^t

To find the roots of a quadratic function we have to use the Bhaskara formula, the roots will give us the time it takes to reach zero height

ax^2 + bx + c = 0

-16^t - 9t + 105 ft = 0 ft

a = - 16 b = - 9 c = 105

t1 = (-b + √ b^2 - 4ac) / 2a

t2 = (-b - √ b^2 - 4ac) / 2a

t1 = (9 + √ (-9^2 - (4 * (-16) * 105))) / 2 * (-16)

t1 = (9 + √ (-81 + 6720)) / - 32

t1 = (9 + √6639) / - 32

t1 = (9 + 81.84) / - 32

t1 = 90.84 / - 32

t1 = - 2.83 s

t2 = (9 - √ (-9^2 - (4 * (-16) * 105))) / 2 * (-16)

t2 = (9 - √ (-81 + 6720)) / - 32

t2 = (9 - √6639) / - 32

t2 = (9 - 81.84) / - 32

t2 = - 72.84 / - 32

t2 = 2.28 s

we have two possible values, we are only going to take the positive one, beacause we are talking about time

t2 = 2.28 s