Which Regions Can This Cannon Reach with Its Projectile?

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The discussion focuses on deriving the trajectory equation for a projectile launched from a cannon at the origin. The correct formula for the projectile's path is identified as y = x tan(α) - (g x²)/(2v₀²)(1 + tan²(α)). Participants clarify the components of the quadratic equation, with a, b, and c defined in relation to the projectile's motion. There is confusion regarding the inclusion of terms in the equation, specifically a term added by Jaan Kalda. The conversation emphasizes the importance of understanding both the horizontal and vertical reach of the projectile in two-dimensional space.
roborangers
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Homework Statement
A cannon is situated in the origin of coordinate axes
and can give initial velocity v0 to a projectile, the shooting direction can be chosen at will. What is the region of space R
that the projectile can reach?
Relevant Equations
but when i checked the solution i say that kalda added y+gx^2/2v_0^2 but i dont understand why
what i tried to do is to write y=v_0tsin alpha - 1/2gt^2 and x=v_0 cos alpha tand that t=x/v_0 cos alphai plug t in the formula for y and get that y= x tan alpha - gx^2/v_0^2 (tan^2 alpha -1)since jaan klada said there should be a quadratic equation (because its a parabola) i thought that gx^2/v_0^2 tan^2 alpha is a, -x tan alpha is b and gx^2/2v_0 is c and got another formula
 
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roborangers said:
Homework Statement: A cannon is situated in the origin of coordinate axes
and can give initial velocity v0 to a projectile, the shooting direction can be chosen at will. What is the region of space R
that the projectile can reach?
Relevant Equations: but when i checked the solution i say that kalda added y+gx^2/2v_0^2 but i dont understand why

what i tried to do is to write y=v_0tsin alpha - 1/2gt^2 and x=v_0 cos alpha tand that t=x/v_0 cos alphai plug t in the formula for y and get that y= x tan alpha - gx^2/v_0^2 (tan^2 alpha -1)since jaan klada said there should be a quadratic equation (because its a parabola) i thought that gx^2/v_0^2 tan^2 alpha is a, -x tan alpha is b and gx^2/2v_0 is c and got another formula
This is not easy to read. Punctuation and spacing are important.
 
PeroK said:
This is not easy to read. Punctuation and spacing are important.
yes you are righ but i got it
 
The correct equation for the projectile trajectory is $$y=x\tan\alpha-\frac{gx^2}{2g}(1+\tan^2\alpha).$$The general equation for the quadratic equation is $$ax^2+bx+c=0$$.What exactly is your question? When you say "What is the region of space R that the projectile can reach?" do you mean in the horizontal direction only or in two dimensional space?

I don't know who Jaan Kalda is, but I think that you should include the whole answer that he provided not just the term that he added.
 
yes exactly i got that y is v_0^2/2g - gx^2/2v_0^2
 
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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