So here is the question my professor made up. I asked if the time was missing and once he thought about it, he said we should be able to figure it out. I need help.
A cannon is mounted on a railroad track on a cliff above the ocean. 300 m below, and standing off a 15 km, is a war ship.The cannon's elevation is fixed and designed to provide maximum range (which I know to be 45°).
a) What is the required muzzle velocity of the cannon?
b) The ship moves with it's range, (It moves to 5km from cliff), so the cannon is rolled back on the track and the projectile charge changed, doubling the cannon's muzzle velocity. How far away does the cannon need to be rolled back to target the ship?
Xf, Yf =X,Y final
Xi, Yi =X,Y initial
(Vx)i, (Vy)i= initial velocity of X, Y
(Vx)f, (Vy)f =final velosity of X,Y
g=gravity of 9.8 m/s^2
1) Xf=Xi+(Vx)i Δt
2) (Vx)f=(Vx)i= constant
The Attempt at a Solution
So first I drew out the problem, I have the cannon on top of the 300m (which equals 3km) cliff and the ship is 15 km out. Then at the muzzle of the cannon I put an angle of 45°. There will be an arch for the cannonball to hit the ship.
(Vx)f and (Vy)f=0
Δt=? (Initial time is 0)
(Vx)i or (Vf)i=?
So I need either the initial velocity of X or Y to solve for Δt, or I really need a Δt. I have no idea how for find the Δt with what I know. Since the arch of the cannonball is not a perfect triangle can I use Ax=Acosθ and Ay=Asinθ? If I can get help on the time then I know I can probably solve A.
Once I have the time I would:
-solve for (Vx)i using eqn. 1.
-Solve for (Vy)i using eqn. 4.
-Then I would add them both to get the muzzle velocity of the cannon
So the muzzle velocity from part A would be doubled
The ship would be 5km from the cliff at this point.
So now I need to find the Xi, so how should I do that exactly?
Thank you so much, I tried to be as detailed as possible and that is all the information that was given for the question.