Cornell Notes — Projectile Motion Subtopic — Notes

Cue / Question	Notes
What stays constant in projectile motion?	Horizontal velocity $u_x = u\cos\theta$ — no horizontal force
What changes in projectile motion?	Vertical velocity: $v_y = u\sin\theta - gt$
At highest point, what is vertical velocity?	$v_y = 0$ ; only $u\cos\theta$ remains
If θ = 45°, what is range?	$R_{max} = \frac{u^2}{g}$ (maximum possible range)
Derive time to reach highest point	$v_y = 0 \Rightarrow t_{up} = \frac{u\sin\theta}{g} = \frac{T}{2}$
What is the trajectory equation?	$y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$ (parabola)
For θ = 30° and θ = 60° same u, compare H	$H_{60} = 3 \times H_{30}$ ; ratio = t $an^{2}$ 60°/t $an^{2}$ 30° = 3
What is the angle for range = height?	Solve: $\frac{u^2\sin 2\theta}{g} = \frac{u^2\sin^2\theta}{2g}$ → θ = 76°

Summary: Projectile motion is two independent motions superimposed. The horizontal is uniform (constant $u\cos\theta$ ); the vertical is uniformly accelerated under g. The trajectory is parabolic. Maximum range at 45°; complementary angles give equal ranges but the larger angle gives greater height.