Unit 2 - Problem 12 - Minimum Angle to Hit Ship

We know the shape of the flight of a cannon ball is a parabola and the vertex form of parabola equation is ( Source = Math Warehouse ):

• $$y = -a(x-h)^2 + k$$

We know three cases so we can  solve for the three unknown variables a, h, and k.:

• $$x= 0 , y=0$$
• $$x=100, y=40$$
• $$x=200, y=0$$

We can substitute our knowns:

• $$0 = -a*h^2 + k$$
• $$40 = -a*(100-h)^2 + k$$
• $$0 = -a*(200-h)^2 + k$$

Solve for definition of k:

• $$k = a*h^2$$
• $$k = -.004 * 100^2$$
• $$k = -.004 * 10000$$
• $$k = 40.0$$

Substitute our knowns:

• $$40 = -a*(110-h)^2 + a*h^2$$
• $$0 = -a*(200-h)^2 + a*h^2$$

Solve for h:

• $$0 = -a((200-h)^2 - h^2)$$
  • $$0 = (200-h)^2 – h^2$$
  • $$h^2 = (200-h)^2$$
  • $$h = 200-h$$
  • $$2h = 200$$
  • $$h = 100$$

Solve for a:

• $$40 = -a((100-h)^2 – h^2)$$
  • $$40/((100-h)^2 – h^2) = -a$$
  • $$a = -40/((100-h)^2 – h^2)$$
  • $$a = -40/((100-100)^2 - h^2)$$
  • $$a = -40/(-10000)$$
  • $$a = -40/(-10000) = 4/10000$$
  • $$a = .004$$

My parabola's equation is:

• $$y = -.004(x-100)^2+40$$

To get initial velocity we need an initial angle.So let's calculate where y is when x = 1 meter.

$$y = -.004(1-100)^2 +40$$

$$y = -.004(9801) + 40$$

$$y = -39.204 + 40 = 0.796$$

$$opp/adj = 0.796/1 = \tan{angle}$$

$$arctan (0.796) = 38.5197895 degrees$$

We know: V² = V0² + 2aΔx

• V = 0 at peak of parabola
• a = 10
• x = 100
• 0 = V0² + 2*10*100
• VO^2 = 2000
• VO = 44.72

Unfortunately the iceberg is between  110 m (as given by the problem) and 115 m (per previous calculations) from iceberg and not at center of parabola (100m). Somehow a higher initial velocity is required to move the parabolic center to the 40 m level of the iceberg.

I don't know how to calculate this problem from here and will do it by trial and error observation.

My best answer is 38.91553 degrees at 45 meters/sec. You can see this is slightly higher values for both initial angle and velocity.

With a real cannon, the initial velocity would be determined by the gunpowder load that would only come in fixed increments. But that would be a different problem.