Question

someone please help !!! =/

A baseball player hits a ball toward the outfield. The height h of the ball in feet is modeled by h(t) = -16t2 + 22t + 3, where t is the time in seconds. If no one catches the ball, how long will it stay in the air? (Round to the nearest tenth of a second and enter only the number.)

HINTS:
When the ball hits the ground, its height is zero, so you are looking for one of the zeros of the quadratic equation.
Though you could use several different methods, the easiest way to solve this particular equation is the quadratic formula (provided here). Take the a, b, a

Answer 1

1) Set h(t) = 0

-16t^2 +22t + 3 = 0.............Multiply both sides of the equation by (-1)

16t^2 - 22t - 3 = 0

Using the quadratic formula,

. . .22 +/- sqrt(484 - (- 192))
t = -------------------------------------
....................32

484 + 192 = 676 = 26^2..............So

t = (-4/32) and ((3/2)

Negative time has no meaning, so the ball is in the air for 1.5 seconds

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2) Since d = 85t and t = 1.5 the distance traveled is 127.5 feet.

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3) Distance = the integral of velocity. The distance from the pitchers rubber to home plate is 60 feet and 6 inches, but, in reality, the ball is released about 59 feet from the plate.

Let h be the horizontal velocity and v the vertical velocity.

h(t) = 116t

∫ 116t dt = 58t^2 = 59

t is approximately 1 (second)

∫ 5 dt = 5t

So the ball started at 6 feet, reached the plate in 1 second still rising and thus was too high to be in the strike zone.

Answer 2

There :)

Answer 3

@Here_to_Help15 thank you!! (:

Answer 4

no problem :)

Answer 5

g2g bye :)