Answer:
The shape of this path is known as a parabola.
Assumptions: there's no air resistance, and that gravitational pull is constant.
Explanation:
The motion of a projectile can be considered in two separate directions:
- In the vertical direction perpendicular to the ground, and
- In the horizontal direction parallel to the ground.
In the vertical direction, the projectile would accelerating downwards at a constant rate. (That rate is equal to [tex]g =\rm 9.81\;m\cdot s^{-2}[/tex] near the surface of the earth.) If [tex]v_{y,0}[/tex] is the initial velocity of the projectile in the vertical direction, the height of the projectile at time [tex]t[/tex] would equal
[tex]\displaystyle y(t) = \frac{1}{2} g\cdot t^{2} + v_{y,0} \cdot t[/tex].
In the horizontal direction, the rocket travels at a constant speed. If [tex]v_x[/tex] is the initial horizontal velocity of the rocket, the horizontal position of the rocket at time [tex]t[/tex] would be
[tex]x(t) = v_x \cdot t[/tex].
Keep in mind that the two-dimensional path of the rocket is more like a function of height [tex]y[/tex] over horizontal position [tex]x[/tex], rather than a function of height [tex]y[/tex] at time [tex]t[/tex]. The goal is to find [tex]y(x)[/tex], an expression of height [tex]y[/tex] in terms of horizontal position [tex]x[/tex].
The relationship between [tex]y[/tex] and [tex]t[/tex] was already determined. Try expressing [tex]t[/tex] using [tex]x[/tex] with the help of the second equation.
[tex]x = v_x \cdot t[/tex].
[tex]\displaystyle t = \frac{x}{v_x}[/tex].
Replace all occurances of the variable [tex]t[/tex] in the expression of [tex]y[/tex] using [tex]\displaystyle \frac{x}{v_x}[/tex].
[tex]\begin{aligned} y &= \frac{1}{2}\cdot \left(\frac{x}{v_x}\right)^{2} + v_{y, 0} \cdot \frac{x}{v_x}\\&= \frac{1}{2v_x}\cdot x^{2} + \frac{v_{y, 0}}{v_x}\cdot x\end{aligned}[/tex].
