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. 3.6-9.8t = 0 t = 18/49 ≈.367 sec y = 3.6(18/49)-4.9(18/49)2+11.4 y ≈ 12.061 meters

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 ≈ 1.936 sec The other answer is negative, so we can ignore it.

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 + (3.6-9.8t)2 dt 1.93626 Time it takes to enter the water from part (b) Length ≈ 12.946 meters

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 -[3.6- 9.8(1.9362)] Θ Tan Θ = .8 -[3.6-9.8(1.9362)] We multiply by -1 because the diver is travelling at negative velocity Θ = 0.051986 rad Θanswer = Π/2 - 0.051986 = 1.5188 rad