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Since the position vector is quadratic so it will be a parabolic path.haruspex said:The acceleration is constant. What familiar path results from constant acceleration?
What planet are you on?rudransh verma said:Idk about constant acceleration path.
Not linearharuspex said:What planet are you on?
In this case: is the position vector quadratic?rudransh verma said:Since the position vector is quadratic so it will be a parabolic path.
Idk about constant acceleration path.
Still working on latexharuspex said:Sure.
To get the scalar eqns from the vector, take the dot product with, respectively, ##\hat x, \hat y##.
Conversely, multiply the scalar equns by respectively, ##\hat x, \hat y##, and add them to produce the vector equn.
Yes.rudransh verma said:Still working on latex