Which Regions Can This Cannon Reach with Its Projectile?

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SUMMARY

The discussion focuses on deriving the projectile motion equation for a cannon situated at the origin of coordinate axes. The correct trajectory equation is established as y = x tan(α) - (g x²)/(2 v₀²)(1 + tan²(α)). Participants clarify the components of the quadratic equation, identifying a = -(g x²)/(2 v₀²), b = -x tan(α), and c = (v₀²)/(2g). The conversation emphasizes the importance of understanding the full equation to determine the region of space R that the projectile can reach.

PREREQUISITES
  • Understanding of projectile motion equations
  • Familiarity with quadratic equations
  • Knowledge of trigonometric functions, specifically tangent
  • Basic physics concepts including gravity (g) and initial velocity (v₀)
NEXT STEPS
  • Study the derivation of projectile motion equations in physics
  • Learn about the implications of varying the angle of launch (α) on projectile range
  • Explore the concept of parabolic trajectories in two-dimensional motion
  • Investigate the effects of initial velocity (v₀) on the maximum height and distance of projectiles
USEFUL FOR

Students and educators in physics, engineers working with projectile dynamics, and anyone interested in the mathematical modeling of motion in two dimensions.

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.
 
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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
 

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