The object will fall back to the ground when s(t) = 0.
So, we need to solve the equation -16t^2 + 11t + 25 = 0 for t.
Using the quadratic formula, we have:
t = (-11 ± sqrt(11^2 - 4*(-16)*25)) / (2*(-16))
t = (-11 ± sqrt(121 + 1600)) / -32
t = (-11 ± sqrt(1721)) / -32
t = (-11 ± 41.5) / -32
So, the possible values for t are:
t = (30.5) / -32 = -0.95
t = (-52.5) / -32 = 1.64
Therefore, the object will fall back to the ground after -0.95 seconds, so the answer is:
E. -0.95
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
A. 0.95
B. −1.64
C. 1.64
D. −0.95
seconds because blank cannot be a solution.
E. -0.95
F. -1.64
G. 1.64
H. 0.95
there are two parts to the question
5 answers
your wrong its 1.64
Apologies for the mistake in my previous response. You are correct that the object will fall back to the ground after 1.64 seconds. Thank you for pointing out the error. The correct answer is:
G. 1.64
G. 1.64
because -1.64 is not a solution right
That is correct. In the context of this problem, a negative time value does not make physical sense as time cannot be negative in this scenario. Therefore, the only valid solution for when the object falls back to the ground is 1.64 seconds. Thank you for clarifying.