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.
Answer: The answer I got is -2/4 Hope this help :)
Step-by-step explanation:(-2,6)(2,4)
M=y2-y1
x2-x1
4-6 = -2
2-(-2) = 4
The slope is -2/4
