Ted popped a baseball straight up with an initial velocity of 48 ft/s. The height, h, in feet, of the ball above the ground is modeled by the function h(t)=-16t2 +48t+3. How long was the ball in the air if the catcher the ball 3 feet above the ground?

3 seconds

Step-by-step explanation:

We have to give us the following function h (t) = - 16 * t ^ 2 + 48 * t + 3

Now, they tell us that the height is 3 feet, therefore replacing that value we have left:

3 = -16 * t ^ 2 + 48 * t + 3, operating cancels 3. And it would be:

-16 * t ^ 2 + 48 * t = 0

Removing common factor -16 * t, it would be:

(-16 * t) * (t - 3) = 0

We have two solutions:

-16 * t = 0 ---> t = 0

 t - 3 = 0 ---> t = 3

Because the time cannot be 0 seconds, the time it would take the ball in the air is 3 seconds.

where are the graphs?

step-by-step explanation: