Question

Definition: -This is the shape of the two-dimensional path followed by a projectile.

Example: This is seen in basketball shots, cannonball firings, and rocket launchings.

Hint: pa

Answer 1

The shape of this path is known as a parabola.

Assumptions: there's no air resistance, and that gravitational pull is constant.

Step-by-step 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
`g =\rm 9.81\;m\cdot s^(-2)` near the surface of the earth.) If
`v_(y,0)` is the initial velocity of the projectile in the vertical direction, the height of the projectile at time
`t` would equal

`\displaystyle y(t) = (1)/(2) g\cdot t^(2) + v_(y,0) \cdot t`.

In the horizontal direction, the rocket travels at a constant speed. If
`v_x` is the initial horizontal velocity of the rocket, the horizontal position of the rocket at time
`t` would be

`x(t) = v_x \cdot t`.

Keep in mind that the two-dimensional path of the rocket is more like a function of height
`y` over horizontal position
`x`, rather than a function of height
`y` at time
`t`. The goal is to find
`y(x)`, an expression of height
`y` in terms of horizontal position
`x`.

The relationship between
`y` and
`t` was already determined. Try expressing
`t` using
`x` with the help of the second equation.

`x = v_x \cdot t`.

`\displaystyle t = (x)/(v_x)`.

Replace all occurances of the variable
`t` in the expression of
`y` using
`\displaystyle (x)/(v_x)`.

`\begin{aligned} y &= (1)/(2)\cdot \left((x)/(v_x)\right)^(2) + v_(y, 0) \cdot (x)/(v_x)\\&= (1)/(2v_x)\cdot x^(2) + (v_(y, 0))/(v_x)\cdot x\end{aligned}`.