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2D Projectile motion formula

by acegikmoqsuwy
Tags: formula, motion, projectile
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Aug9-14, 11:26 AM
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 [itex]V_n[/itex]. Let the terminal side of the launch angle with respect to the ground be denoted [itex]\theta[/itex]. The vertical component is [itex]V_y[/itex] and the horizontal component is [itex]V_x[/itex]. The acceleration of gravity is denoted [itex]g[/itex]. The time is [itex]t[/itex], and the height of the projectile at a particular time is [itex]h[/itex]. Finally, the displacement from the launch point to the point where the projectile hits the ground is denoted [itex]s[/itex].

We see that [itex]V_y = V_n \sin (\theta)[/itex] and [itex]V_x = V_n \cos (\theta)[/itex]. Since the displacement is strictly horizontal, [itex]s = V_x t[/itex].

We can model the height using the formula for displacement with an initial velocity and constant acceleration [itex]\displaystyle h = V_y t - \frac12 g t^2[/itex].

Solving for the time that we need (when the height becomes zero), we get [itex]\displaystyle V_y t - \frac12 g t^2=0[/itex] so [itex]\displaystyle t \left(V_y - \frac12 g t\right) = 0[/itex].

This means [itex]\displaystyle t = \frac {2 V_y} g[/itex]. Thus [itex]\displaystyle s = \frac {2 V_y V_x} {g}[/itex].

Substituting for the values of [itex]V_y[/itex] and [itex]V_x[/itex] we get [itex]\displaystyle s = \frac {2 V_n^2 \sin (\theta) \cos (\theta)} {g} = \frac {V_n^2} {g} \sin (2 \theta)[/itex].
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Doc Al
Aug9-14, 11:31 AM
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P: 41,579
Looks good to me.

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