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?

28.5 ft

Step-by-step explanation:

The height of a ball is given by the equation ƒ (x) = - 16x² + 45x + 2.5 where x is the tome the ball travels. To find the height of the ball after a seconds, we substitute a in place of x, the equation becomes:

ƒ (a) = - 16a² + 45a + 2.5.

Therefore if the height of the ball is needed to be found after 2 seconds, we find f (2).

ƒ (2) = - 16 (2) ² + 45 (2) + 2.5 = - 64 + 90 + 2.5 = 28.5 ft

f (2) = 28.5 ft