SOLUTION: Given s(t) = -16t^2 + 32t + 48, where t = 0 represents the time that object is released (thrown upward). 1.1 What is s(o) and what does it represent? 1.2 What is the value of s w

Question 947834: Given s(t) = -16t^2 + 32t + 48, where t = 0 represents the time that object is released (thrown upward).
1.1 What is s(o) and what does it represent?
1.2 What is the value of s when the object hits the ground?
How many seconds (from time of release) will it take for the object to hit the ground?
1.3 What is the relevant domain for the model? (what values for t make sense?)
1.4 What does a sketch of y = s(t) look like?
1.5 At what time does the object attain a maximum height? What is the maximum height?
Than you in advance.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
1.1) s(0) = 48, s(0) represents the height above the ground when the object is released
1.2) s is 0
0 = -16t^2 + 32t + 48
using quadratic formula to solve equation, we have
t = -1 or t = 3, we want the positive value t = 3
which means the object will hit the ground in 3 seconds
1.3) 0 < or = t < or = 3
1.4) graph of -16t^2 + 32t + 48

1.5) set t = the axis of symmetry
t = -b/2a = -32 / (2*(-16)) = 1, max height at 1 second
s(1) = -16 +32 +48 = 64, max height is 64

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