An object is launched vertically upward from a 25 foot platform and is modeled by the function s(t)=−16t2+11t+25.
After how many seconds will the object fall back to the ground?
seconds because
cannot be a solution.
(Hint: Does not factor, use different method to solve)
A. -1.64
B.-0.95
c.0.95
d.1.64
Given the context of the problem and the physical situation of an object being launched vertically upward, a negative time value does not make sense, as time cannot be negative in this scenario.
Therefore, the negative solution of approximately -1.64 seconds is not a valid solution and should be ignored.
The object will fall back to the ground after approximately 1.64 seconds. So, the correct answer is:
D. 1.64
To find when the object will fall back to the ground, we need to find the time when s(t) = 0.
0 = -16t^2 + 11t + 25
This is a quadratic equation that we can solve using the quadratic formula:
t = (-11 ± sqrt(11^2 - 4(-16)(25))) / (2(-16))
t = (-11 ± sqrt(121 + 1600)) / -32
t = (-11 ± sqrt(1721)) / -32
Since we cannot have a negative time for this problem, we can ignore the negative solution and use only the positive solution:
t = (-11 + sqrt(1721)) / -32 ≈ 1.64
Therefore, the object will fall back to the ground after approximately 1.64 seconds. So, the correct answer is:
D. 1.64
seconds because
cannot be a solution.
Since we cannot have a negative time for this problem, we can ignore the negative solution and use only the positive solution.
Therefore, the object will fall back to the ground after approximately 1.64 seconds. So, the correct answer is:
D. 1.64