A ball is thrown from a height of 139 feet with an initial downward velocity of 15 ft/s. The ball's height h (in feet) after t seconds is given by the following.

h=139-15t-16t^2

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.)

t = _______ seconds???

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wegnerkolmp2741o wegnerkolmp2741o · Answer 1 · 2019-02-03T22:54:28+01:00

Answer:

t = 2.52 seconds

Step-by-step explanation:

h=139-15t-16t^2

We want to know when the ball hits the ground

That would be when h=0

0 = 139-15t-16t^2

We can use the quadratic formula to find t

t = -b ± sqrt(b^2-4ac)

----------------------

2a

where a = -16 b = -15 and c = 139

t = -(-15) ± sqrt((-15)^2-4(-16)139)

----------------------

2(-16)

t = (15) ± sqrt(225+8896)

----------------------

-32

t = (15) ± sqrt(9121)

----------------------

-32

t = 15+ sqrt(9121) t = 15- sqrt(9121)

-------------------- or -------------------

-32 -32

-3.453247707 or 2.515747707

Since time cannot be negative

2.515747707

Round to the nearest hundredth

t = 2.52 seconds