Question

If an object is launched vertically upward from a 25 foot platform and is modeled by the function s(t)=-16+^2+11t+25, the object will fall back to the ground after ______, select the correct answer from the following: a) 1.64 b) -1.64 c) 0.95 d) -0.95; seconds because _____ cannot be a solution: Select the correct answer from the following a) 1.64 b) 0.95 c) -1.64 d) -0.95

Answer 1

So, are you saying the object will fall back to ground after 1.64 seconds, and in answering the SECOND PART of the question are you saying 1.64 cannot be a solution?

Answer 2

Are you saying the correct answer for SECOND PART of the question is -1.64?

Answer 3

If the object falls back to the ground, it means that the height is zero (s(t) = 0).

To find when this happens, we need to solve the equation:
0 = -16t^2 + 11t + 25

We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula.

Using the quadratic formula, we have:
t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -16, b = 11, and c = 25.

t = (-11 ± √(11^2 - 4(-16)(25))) / (2(-16))
t = (-11 ± √(121 + 1600)) / (-32)
t = (-11 ± √(1721)) / (-32)

The square root of 1721 is approximately 41.48.

t = (-11 ± 41.48) / (-32)

Using the positive square root:
t = (-11 + 41.48) / (-32)
t = 30.48 / (-32)
t ≈ -0.95

Using the negative square root:
t = (-11 - 41.48) / (-32)
t = -52.48 / (-32)
t ≈ 1.64

Therefore, the object will fall back to the ground after approximately 1.64 seconds.

Now, to determine which time cannot be a solution, we can look at the equation s(t) = -16t^2 + 11t + 25.

The height of the object cannot be negative, so we are looking for the time when the height is 0.

At t = -0.95, s(-0.95) = -16(-0.95)^2 + 11(-0.95) + 25 ≈ 25.375
At t = 1.64, s(1.64) = -16(1.64)^2 + 11(1.64) + 25 ≈ 25.375

Since both values result in a positive height, none of the given options (a) 1.64, (b) 0.95, (c) -1.64, (d) -0.95 cannot be a solution.

Therefore, the correct answer is: The object will fall back to the ground after approximately 1.64 seconds because none of the given options can be a solution.

Answer 4

Apologies for the confusion in my previous response. Let me clarify:

The correct answer is:
The object will fall back to the ground after approximately 1.64 seconds, and 1.64 cannot be a solution.

To solve the quadratic equation -16t^2 + 11t + 25 = 0, we find two solutions using the quadratic formula:

t = (-11 ± √(11^2 - 4(-16)(25))) / (2(-16))
t ≈ -0.95 and t ≈ 1.64

So, the object will fall back to the ground after approximately 1.64 seconds. Among the given options, 1.64 cannot be a solution because it is a valid solution to the equation. Therefore, the correct answer is "c) -1.64" as none of the other options can be a solution.

Answer 5

Apologies for the confusion. Let me clarify:

The correct answer for the second part of the question is "b) 0.95".

When solving the quadratic equation -16t^2 + 11t + 25 = 0, we find two solutions using the quadratic formula:

t = (-11 ± √(11^2 - 4(-16)(25))) / (2(-16))
t ≈ -0.95 and t ≈ 1.64

In this case, we are looking for the time when the object falls back to the ground, which means the height is 0. Evaluating the function s(t) = -16t^2 + 11t + 25 at these times:

s(-0.95) = -16(-0.95)^2 + 11(-0.95) + 25 ≈ 0.95
s(1.64) = -16(1.64)^2 + 11(1.64) + 25 ≈ 0.95

Both s(-0.95) and s(1.64) are approximately equal to 0. Therefore, neither -0.95 nor 1.64 can be a solution.

Hence, the correct answer is "b) 0.95" as it cannot be a solution.