Which Regions Can This Cannon Reach with Its Projectile?

AI Thread Summary
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
Messages
3
Reaction score
1
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
 
Physics news on Phys.org
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
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...

Similar threads

Back
Top