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RideAnS [48]
3 years ago
11

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?
Mathematics
2 answers:
Nesterboy [21]3 years ago
8 0
4=-16t^2+19t+5 
<span>16t^2-19t-1=0 </span>
<span>quadform is (-b+-sqrt(b^2-4ac))/2/a </span>

<span>a=16 </span>
<span>b=-19 </span>
<span>c=-1 
</span>Plug in to the quadratic formula.
<span>1.238 and -.0505.
cant be negative, so it has to be 1.238.</span>
scoray [572]3 years ago
4 0
-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.

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Answer and explanation:

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