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A ball is thrown from a height of 49 meters with an initial downward velocity of 2 m/s. The ball's height (in meters) after t seconds is given by the following.

h = 49 - 2t -[tex]5t^{2}[/tex]

How long after the ball is thrown does it hit the ground?
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)

A ball is thrown from a height of 49 meters with an initial downward velocity of 2 ms The balls height in meters after t seconds is given by the following h 49 class=

Respuesta :

musiclover10045

The height of the ball at ground level would be 0 feet.

Set h to 0 and solve for t:

Rewrite the equation as  

-5t^2 -2t +49 = 0

Factor -1 out of the equation to get:

5t^2 + 2t -49 = 0

Use the quadratic formula:

-b +/- sqrt(b^2 -4(ac) / 2a

Where a = 5, b = 2 and c = -49

The quadratic formula becomes:

-2 +/-sqrt(2^2 – 4(5*-49) / (2*5)

Simplify to get :

T = -2 +/-sqrt246 / 10

T = 2.936 and – 3.336

Since the time has to be a positive value t = 2.936 seconds.

Rounded to the nearest hundredth = 2.94 seconds.

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