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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?

User Lumi Lu
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2 Answers

5 votes
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.

User Papey
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8.0k points
6 votes
-16 t² + 19 t + 5 = 4
-16t² + 19 t + 1 = 0

t_(12) = \frac{-19\pm \sqrt{19^(2)-4*(-16)* 1 } }{-32}= \\ (-19\pm √(361+64) )/(-32) = (-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.

User Greg Grater
by
8.2k points
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