Question

At the beginning of a basketball game, the referee tosses the ball vertically into the air. Its height, h, in feet after t seconds

is given by h(t) = -16t² + 24t +5. During what time interval (to the nearest tenth of a second) is the height of the ball
greater than 9 feet?

Answer 1

Explanation:

so, we need to find the points of time, when the bank is exacting 9 ft in the air. the interval is between these 2 points.

9 = -16t² + 24t + 5

-16t² + 24t - 4 = 0

we can simplify

-4t² + 6t - 1 = 0

the general solution for a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = t

a = -4

b = 6

c = -1

t = (-6 ± sqrt(36 - 4×-4×-1))/(2×-4) =

= (-6 ± sqrt(36 - 16))/-8 = (-6 ± sqrt(20))/-8

t1 = (-6 + sqrt(20))/-8 = 6/8 - sqrt(20)/8 =

= 3/4 - 2×sqrt(5)/8 = 3/4 - sqrt(5)/4 = 0.190983006... s

t2 = (-6 - sqrt(20))/-8 = 6/8 + sqrt(20)/8 =

= 3/4 + 2×sqrt(5)/8 = 3/4 + sqrt(5)/4 = 1.309016994... s

so the ball is higher than 9 ft between 0.2 and 1.3 seconds.