Question

An outfielder throws a baseball to the player on third base. The height h of the ball in feet is modeled by the function h(t) = -16t^2 + 19t + 5, where t is time in seconds. The third baseman catches the ball when it is 4 feet above the ground. To the nearest tenth of a second, how long was the ball in the air before it was caught?

Answer 1

H(t) = -16t² + 19t + 5

Since you are given H in feet, just plug the number into the equation and solve.

4t = -16t² + 19 t + 5

0 = -16t² + 19t + 5 - 4t
0 = -16t² + 15t + 5

5 = -16t² +15t
16t² - 15t= -5

t(16t - 15) = -5

t ≠ -5 because the ball can not be under the ground
so you are left with 16t - 15 = -5

16t = 10
t = 10/16

Simplify
5/8 in decimal form that is
0.625s