1)
An object is launched vertically upward from a 25 foot platform and is modeled by the function s (t) = -16t^2 + 11t + 25
The object will fall back to the ground after (1 point)
answer is 0.95
seconds because ______ cannot be a solution.
A -0.95
B -1.64
C 0.95
D 1.64
The object cannot fall back to the ground in negative time, so we can eliminate option A (-0.95) and option B (-1.64).
To find the time it takes for the object to fall back to the ground, we set the function s(t) equal to 0 (since the object is at the ground).
Therefore, we have:
-16t^2 + 11t + 25 = 0
Using the quadratic formula, we can find the roots of this equation:
t = (-11 ± √(11^2 - 4*(-16)*25)) / (2*(-16))
t = (-11 ± √(121 + 1600)) / (-32)
t = (-11 ± √1721) / (-32)
t ≈ (-11 ± 41.48) / (-32)
Therefore, the possible solutions are:
t ≈ (30.48) / (-32) ≈ -0.95
t ≈ (-52.48) / (-32) ≈ 1.64
Since time cannot be negative for this situation, we can conclude that the object will fall back to the ground after approximately 1.64 seconds.
Therefore, the answer is option D 1.64.