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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).
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).
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