1. Part (a) Integrate dy/dt to get the vertical distance.
We need to maximize vertical distance so we set dy/dt equal to 0
y = 3.6t-4.9t2+11.4
We know c = 11.4 because the diver starts 11.4 m above the water.
t = 0
t = 18/49 ≈.367 sec
y = 3.6(18/49)-4.9(18/49)2+11.4
y ≈ meters

2. We want y = 0, therefore we set the y equation from part a equal to 0.
Part (b)
We want y = 0, therefore we set the y equation from part a equal to 0.
0 = 3.6t-4.9t2+11.4
A ≈ sec
The other answer is negative, so we can ignore it.

3. We need to find curve length!!
Part (c)
We need to find curve length!! a
Remember:
Length = ∫ √(dx/dt)2 + (dy/dt)2 dt b
Length = ∫ √(.8)2 + ( t)2 dt
Time it takes to enter the water from part (b)
Length ≈ meters

4. We multiply by -1 because the diver is travelling at negative velocity
Part (d)
This is a vector problem, so we will use the given derivative equations to fill in the triangle.
0.8
-[ (1.9362)] Θ
Tan Θ = .8
-[ (1.9362)]
We multiply by -1 because the diver is travelling at negative velocity
Θ = rad
Θanswer = Π/
= rad