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A ball is thrown downward from the top of a 200-foot building with an initial velocity of 20 feet per second. The height of the ball h in feet after t

seconds is given by the equation h= - 16t-20t+200. How long after the ball is thrown will it strike the ground?
The time it takes for the ball to strike the ground is about
(Simplify your answers. Type an integer or decimal rounded to the nearest tenth as needed.)

1 Answer

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Answer: When the ball strikes the ground, its height above the ground is zero. So we can set h = 0 in the equation and solve for t:

0 = -16t^2 - 20t + 200

Dividing both sides by -4 (to simplify the equation):

0 = 4t^2 + 5t - 50

Now, we can use the quadratic formula to solve for t:

t = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 4, b = 5, and c = -50.

t = (-5 ± sqrt(5^2 - 4(4)(-50))) / 2(4)

t = (-5 ± sqrt(625)) / 8

t = (-5 ± 25) / 8

So the two possible solutions are:

t = 3/2 or t = -5

Since time cannot be negative, the only valid solution is t = 3/2.

Therefore, the ball will strike the ground 1.5 seconds after it is thrown.

Explanation:

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