StarAce2017
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A ball is thrown from a height of 80 feet with an initial downward velocity of 4/fts. The ball's height h (in feet) after t seconds is given by the following.
h=80-4t-16t^2
How long after the ball is thrown does it hit the ground? Round your answer(s) to the nearest hundredth.

Respuesta :

SouthernRider

Answer:

2.11 sec

Step-by-step explanation:

Given that the ball is thrown from a height of 80 feet

it follows the equation

[tex]h = 80-4t-16t^2[/tex]

when the ball hits the ground the height if the ball is equal to 0

[tex]16t^2+4t-80=0[/tex]

[tex]t=\frac{-4+\sqrt{4^2-4\times 16\times(-80)} }{2\times 16}[/tex]

[tex]t=\frac{-4+71.66}{32}[/tex]

[tex]t=2.11[/tex]

0700675

Alright, so we have h=80-4t-16t^2. We have to find when h=0. We can use the quadratic formula

(use x as t)

x=(4+-sqrt(16-(-64*80))/32=(4+-sqrt(5136))/(-32) , and since 4+sqrt(5136)/(-32) is clearly negative that doesn't work, so we have (4-sqrt(5136))/(-32), which is approximately 2.11

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