To begin a basketball game, a referee msut toss the ball vertically into the air. This process can be modeled by the equation h(t) = -16t^2+22t+6, where h represents the ball's height (in feet) after t seconds. Determine the time interval (in seconds) in which the height of the basketball is 8 feet. Round your answer to the nearest thousandth of a second

1.277s and 0.098s

Step-by-step explanation:

Given the height of the ball modelled by the equation

h(t) = -16t^2+22t+6, where;

t is the time in seconds

To calculate the time interval (in seconds) in which the height of the basketball is 8 feet, we will simply substitute h(t) = 8 into the expression and calculate t as shown;

8 = -16t^2+22t+6

Divide through by 2

4 = -8t^2+11t+3

-8t^2+11t+3-4 = 0

8t^2-11t+1 = 0

t = 11±√11²-4(8)/2(8)

t = 11±√121-32/16

t = 11±√89/16

t = 11±9.43/16

t = 11+9.43/16 and t = 11-9.43/16

t = 20.43/16 and t = 1.57/16

t = 1.277s and t = 0.098s

Hence the time interval to the nearest thousands are 1.277s and 0.098s

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