A fixed cannon is to fire a projectile at a tank

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The discussion focuses on determining the firing angle for a cannon aimed at a tank moving toward it at 40 mph, starting from a distance of 8 miles. The projectile's muzzle speed is 1000 mph, and the equations of motion are derived using initial conditions and gravitational acceleration. The solution involves relating the projectile's velocity to time and distance in both the x and y directions. A key suggestion is to eliminate the time variable to simplify the equations. The expected outcome of the calculations is a specific equation involving trigonometric functions of the firing angle.
esmmajor
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1. Problem Statement:
A fixed cannon is to fire a projectile at a tank moving toward the cannon at 40 mph. If the cannon is to fire at the moment the tank is 8 miles from the cannon, and the muzzle speed of the projectile is 1000 mph what is the correct equation to determine the firing angle?2. Homework Equations :
Acceleration of gravity: 79000 m/h^23. Solution Attempt:

Initial conditions:
r=<8-40t,0>
v=<1000*cos(theta), 1000*sin(theta)>

started with:
acceleration_projectile=<0,-g>
<0,-79000.
velocity (took integral and used initial conditions to come up with constants)
v=<1000*cos(theta), -79000t+1000*sin(theta)>

position:
r=<1000*cos(theta)*t+t-40t, -79000t^2/2+1000*sin(theta)t>

**The answer I'm supposed to get is (3.16*cos(theta)+0.127)*sin(theta)=1

Any help would be greatly appreciated.
 
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Hi, esmmajor.
Well, you are right for every step. Except that, at some point, you want to eliminate t.
My personal suggestion would be:
in x direction, how would you relate the velocity of the particle to the time and distance?
and in y direction, how are you going to relate the initial velocity to the acceleration and time?
And finally, are these two time t, equal?
Have an idea of how to eliminate the t?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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