Projectile motion, need to find angle

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A cannon with a muzzle speed of 1009 m/s needs to be fired at an angle to hit a target 1900 m away and 795 m above. The initial attempt involved using a projectile motion equation but became complicated due to the angle being inside trigonometric functions. The user then separated the problem into x and y components, deriving equations for both. By expressing the cosine and sine of the angle in terms of time, they aimed to simplify the problem further. Ultimately, the user resolved the issue with assistance, indicating progress in solving for the firing angle.
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Homework Statement


A cannon with a muzzle speed of 1009 m/s is used to start an avalanche on a mountain slope. The target is 1900 m from the cannon horizontally and 795 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.)



Homework Equations


I'm thinking this equation is my best bet, yf = tan(angle)xf - (g/(2 * vi^2 * cos^2(angle))) * (xf^2)


The Attempt at a Solution


since I know the vi, xf, and yf, I was able to plug everything in and reached 795 = 1900tan(angle) - 17.375/cos^2(angle). This is where I am stuck. I tried several times but could not find a way to solve for angle.

I would really appreciate it if someone could point out how to solve for the angle in my problem. Or is there a totally different way of solving this problem?
 
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Your work DOES look great (I checked your first equation). The problem is that you've got an algebraically difficult thing to solve for (the angle) because it's inside trig functions.

What I suggest is keeping your equation in terms of the components. Relate one of these components ONLY to the initial speed and the remaining component. Then solve for the component you kept. You can use this solution to find the angle.

Added at a later edit: You could also just graph your second function and look for where it crosses zero, but that's easy only because of modern graphing tools.
 
Last edited:
thank you for your input, i separated my problem into x and y components:
x component ---> 1900 = 1009cos(theta) * t
y component ---> 795 = 1009sin(theta) * t - 4.9t^2

I used my x component to solve for cosine
1900/1009 = cos(theta) * t
cos(theta) = 1.883/t

I then used trig identity sin^2(theta) + cos^2(theta) = 1
sin^2(theta) + (1.883/t)^2 = 1
sin(theta) = square root(1 - (1.883/t)^2)

then I plugged sin(theta) equation into my y component
795 = 1009 * square root(1-(1.883/t)^2) * t -4.9t^2

I think I'm on the right track but I don't really know where to go with this, it's not really a quadratic equation and I don't know what to do now. Any pointer would be greatly appreciated.
 
nm, I got it. Thank you for all your help.
 
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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