# 2D Projectile motion formula

by acegikmoqsuwy
Tags: formula, motion, projectile
 P: 11 Is this a correct derivation of a relation between displacement of a particle with its terminal launch angle, velocity, and environmental gravity? Let the velocity of the projectile be denoted $V_n$. Let the terminal side of the launch angle with respect to the ground be denoted $\theta$. The vertical component is $V_y$ and the horizontal component is $V_x$. The acceleration of gravity is denoted $g$. The time is $t$, and the height of the projectile at a particular time is $h$. Finally, the displacement from the launch point to the point where the projectile hits the ground is denoted $s$. We see that $V_y = V_n \sin (\theta)$ and $V_x = V_n \cos (\theta)$. Since the displacement is strictly horizontal, $s = V_x t$. We can model the height using the formula for displacement with an initial velocity and constant acceleration $\displaystyle h = V_y t - \frac12 g t^2$. Solving for the time that we need (when the height becomes zero), we get $\displaystyle V_y t - \frac12 g t^2=0$ so $\displaystyle t \left(V_y - \frac12 g t\right) = 0$. This means $\displaystyle t = \frac {2 V_y} g$. Thus $\displaystyle s = \frac {2 V_y V_x} {g}$. Substituting for the values of $V_y$ and $V_x$ we get $\displaystyle s = \frac {2 V_n^2 \sin (\theta) \cos (\theta)} {g} = \frac {V_n^2} {g} \sin (2 \theta)$.