Question

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

Answer 1

Answer:

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