A soccer ball is kicked from the ground, and its height in meters above ground is modeled by the function h(t) = -4.9t2 + 19.6t, where t represents the time in seconds after the ball is kicked. how long is the ball in the air?

Given the quadratic function:

h(t) = -4.9t^2 + 19.6t, where t represent time in seconds/

Determine how long is the ball in the air.

Solution:

Derive the given function  first and equate to 0, then solve for t.

In order to derive a given function, we apply the rules of derivative, in which we multiply the degree with the term, while the degree will be subtracted by 1. Derivative is denoted as y' or for the given function h'(t) read as h prime or derivative of h.

h(t) = -4.9t^2 + 19.6t

h'(t) = -9.8t  + 19.6

We multiplied 2 by -4.9 then we subtract 2 by 1, leaving behind t. For 19.6 since t has a degree of 1, if we multiply it by 19.6 it will still remain, and when we subtract 1 by 1 the answer will be 0 and t^0 = 1 leaving behind only 19.6.

Now we equate h'(t) to 0

-9.8t  + 19.6 = 0

9.8t = 19.6

Divide both sides by 9.8 in order to determine the time.

t = 2

The ball is 2 seconds in the air.