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) = -16t2 + 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

-16 t² + 19 t + 5 = 4
-16t² + 19 t + 1 = 0
$t_{12} = \frac{-19\pm \sqrt{19^{2}-4*(-16)* 1 } }{-32}= \\ \frac{-19\pm \sqrt{361+64} }{-32} = \frac{-19\pm20.61}{-32}$
We will accept: ( another answer is negative )
t = 1.2 s ( to the nearest tenth ) 
Answer: the ball was 1.2 seconds in the air before it was caught.

Answer 2

4=-16t^2+19t+5 
16t^2-19t-1=0 
quadform is (-b+-sqrt(b^2-4ac))/2/a 

a=16 
b=-19 
c=-1 
Plug in to the quadratic formula.
1.238 and -.0505.
cant be negative, so it has to be 1.238.