Answer to Question #102199 in Mechanics | Relativity for Ceaser daniel

Question #102199
The equation of a parabola is y = ax2 + bx + c, where a, b and c are constants. The
x- and y-coordinates of a projectile launched from the origin as a function of time are
given by x = v0xt and y = v0yt − 1gt2, where v0x and v0y are the components of the 2
initial velocity.
(a) Eliminate t from these two equations and show that the path of a projectile is a
parabola and has the form y = ax2 + bx.
(b) What are the values of a, b and c for the projectile?
Expert's answer

From the conditions of the problem we have

"x=v_{0x}\\cdot t" (1)

"y=v_{0y}\\cdot t-g\\cdot t^2" (2)

From (1) we write


substitute in (2)

"y=v_{0y}\\cdot \\frac{x}{v_{0x}}-g\\cdot (\\frac{x}{v_{0x}})^2=x\\cdot \\frac{v_{0y}}{v_{0x}}-x^2\\cdot \\frac{g}{(v_{0x})^2}=-\\frac{g}{(v_{0x})^2}\\cdot x^2+\\frac{v_{0y}}{v_{0x}}\\cdot x" (3)

comparing equation (3) with equation "y=ax^2+bx+c"

get the expression




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