College Algebra

Question

A varsity soccer player kicks a soccer ball approximately straight upward with an initial velocity of 68 ft/sec. The ball leaves the players foot at a height of 3 feet. 

Part 1 of 2: (2 pts)

Use the position function to write a model representing the height of the soccer ball (in feet) t seconds after being kicked. (Assume the acceleration due to gravity is 32 ft/sec²). Be sure to reduce your first term.

Part 2 or 2: (6 pts)

Use that model to determine the times at which the ball is 73 feet in the air. Use FACTORING to solve. Be sure to write a sentence that explains both of your answers.

Answer 1

1)

Position function x(t) = -32t^2 + v0 * t + H0

where V0 and H0 are the initial velocity and height, respectively

So then x(t) = -32t^2 + 68t + 3

2) -32t^2 + 68t + 3 = 73

-32t^2 + 68t - 70 = 0

16t^2 - 34t + 35 = 0

B^2 - 4ac = (-34)^2 - 4(16)(35) = 1156 - 2240 < 0

so this quadratic equation has two imaginary/complex solutions

the reason is because the max occurs at -b/(2a) =

34/32 = 17/16

and x(17/16) = -32(17/16)^2 + 68(17/16) + 3 =

-32(289/256) + 68(17/16) + 3 =

(-32*289)/256 + (68*17*16)/256 + 3 =

9248/256 + 3 =

36 and 32/256 + 3 =

36 and 1/8 + 3 =

39 and 1/8

in other words, the ball does not go that high...

please repost