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

Question

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?

Answer 1

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

$t=\frac{x}{v_{0x}}$

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

$a=-\frac{g}{(v_{0x})^2}$

$b=\frac{v_{0y}}{v_{0x}}$

$c=0$