This must be on the moon as the acceleration due to gravity in this equation must be around 1/8 that on earth. :) Anyway...
h=-2t^2+9t+11
A)
dh/dt=-4t+9, when velocity, dh/dt=0, it is the maximum height reached
dh/dt=0 only when 4t=9, t=2.25 seconds
h(2.25)=21.125 ft (21 1/8 ft)
B)
As seen in A), the time of maximum height was 2.25 seconds after the squirrel jumped.
C)
The squirrel reaches the ground when h=0...
0=-2t^2+9t+11
-2t^2-2t+11t+11=0
-2t(t+1)+11(t+1)=0
(-2t+11)(t+1)=0, since t>0 for this problem...
-2t+11=0
-2t=-11
t=5.5 seconds.
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I hope this helps:)
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first would be 5-6 and second would be 8-10
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i think it's a
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