Question

Derive an equation for the path of a projectile fired parallel to horizontal.

Answer 1

Question 32899

Let the initial y-component be $h$ .

For accelerated motion, velocity is $v = v_{0} + at$ . There is no horizontal acceleration (only vertical acceleration $g = 9.81\frac{m}{s^{2}}$ according to gravity). Initial velocity $v_{0} = v_{0x}$ only has horizontal component.

Therefore, $v_{x} = v_{0}$ , $v_{y} = -gt$ .

Integrating previous two equations, obtain: $x(t) = v_0 t$ ; $y(t) = h - \frac{gt^2}{2}$ .

In order to obtain path in usual form $y = y(x)$ , exclude $t$ from $y(t)$ by expressing it from $x(t)$ ( $t = \frac{x(t)}{v_0}$ ):

$y(x) = h - \frac{gx^2}{2v_0^2}$ - this is the equation for the path of a projectile fired parallel to horizontal.