A ball is kicked from an initial height of 2.5 feet with an initial velocity of 45 feet per second. The function ƒ(x) = −16x^2 + 45x + 2.5 models its
path, where x is the time (in seconds) the ball travels and ƒ(x) is the height (in feet) the ball is kicked. What is the height of the ball 2 seconds after it is kicked?

b

step-by-step explanation:

the image is given below

since the options are not given so we draw it from the equation

the given equation is

y = (x-1) (x+4)

y = y² + 3x - 4

the above equation is a quadratic equation which will have two roots. if we draw the graph for it, it will look like this

60: 45 =12: 9 =4: 3 so the answer is 4: 3

24,18t

step-by-step explanation: