Afootball is kicked into the air from the initial height of 3 ft. the height, in feet, of the football above the ground is given by s(t) = -16t^2 + 35t + 3, where t is time in seconds and t is greater than or equal to 0. which is closest to the time when the football will be 20 ft above ground? a. 2.72 sec b. 0.73 sec or 1.46< c. 0.53 sec d. 0.53 sec or 2.72 sec

B

Step-by-step explanation: As you can see, -16t^2 + 35t + 3 is a quadratic equation, where t is the time and s is the height. Since we have the height and we know that 20 is equal to -16t^2 + 35t + 3 we can solve it.

Let's order it: 20 = -16t^2 +35t + 3 -> 16t^2 - 35t - 3 + 20 = 0 -> 16t^2 - 35t + 17 = 0

And we can use the quadratic formula.

t1,2 = 35±√1225-1088 / 32 = 35±√137 / 32

We have t1 = 35 + √137 / 32 = 0,727 which we can approximate to 0,73

t2 = 35 - √137 / 32 = 1,459 which we can approximate to 1,46

Why do we have two solution? Of course, it is a quadratic equation and it has two solution but expecially because one is the time when the ball goes up and one is the time when it goes down.

i think the amount would be $140.62