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.
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.
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