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a rock is thrown from the top of a tall building. the distance, in feet, between the rock and the ground x seconds after it is thrown is given by f(x)=-16x62-4x+382 How long after the rock is thrown does it hit the ground

User Kound
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2 Answers

5 votes

Final answer:

It takes approximately 5.0 seconds for the rock to hit the ground after it is thrown.

Step-by-step explanation:

To find the time when the rock hits the ground, we need to find the value of x when the distance between the rock and the ground is 0. In the given equation, f(x) = -16x^2 - 4x + 382, where f(x) represents the distance between the rock and the ground. We can set this equation equal to 0 and solve for x:



-16x^2 - 4x + 382 = 0



This is a quadratic equation. We can use the quadratic formula to solve for x:



x = (-b ± sqrt(b^2 - 4ac)) / (2a)



Plugging in the values from the equation, we get:



x = (-(-4) ± sqrt((-4)^2 - 4(-16)(382))) / (2(-16))



This simplifies to:



x = (4 ± sqrt(16 + 24512)) / (-32)



x = (4 ± sqrt(24528)) / (-32)



x ≈ (-4 ± 156.49) / (-32)



Since we're dealing with time, the negative value doesn't make sense in this context. So, we take the positive root:



x ≈ (4 + 156.49) / (-32)



x ≈ 160.49 / (-32)



x ≈ -5.0



Therefore, it takes approximately 5.0 seconds for the rock to hit the ground after it is thrown.

User Ofek Glick
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3 votes
The first thing we are going to do in this case is to rewrite the function correctly.
We have then:
f (x) = - 16x ^ 2-4x + 382
The moment the rock hits the ground we have f (x) = 0
0 = -16x ^ 2-4x + 382
We solve the polynomial looking for its roots which are:
x1 = -5.012803699004288
x2 = 4.762803699004288
We ignore the negative root. The solution is:
x = 4.76
Answer:
it hits the ground 4.76 s after is thrown
User Tinple
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