Unit 2 - Problem 10 - Min Angle To Clear Iceberg

Know by observation an angle of 19.071degrees just clears iceberg.

Given height of iceberg is 40m.

Iceberg Height = Opposite

Distance to the iceberg = Adjacent

Opposite/Adjacent = tan(angle to the top of iceberg)

40/distance = sine(19.071 degrees)

distance = 40/tan(19.071 degrees)

40/tan(19.071 degrees) = 115.702472m to iceberg peak

Calculate the vertical velocity it takes to reach just 40m vertical

V² = V0² + 2aΔx

0 = (Vup)^2 + 2*(-10)*115.702472m

0 = (Vup)^2 – 2314.049544

(Vup)^2 = 2314.049544

Vup = 48.1045688474

Calculate the total initial velocity to get 40m up and 115.702472m out

Vup/Vtotal = Opposite/Adjacent = sin(19.071) = 0.326739575

Vtotal = Vup/sin(19.071)

Vtotal = 48.1045688474 / 0.326739575

Vtotal = 147.226025

This will overshoot the ship we are aiming at.