NoteTube

Part of ME-02 — Kinematics

Cornell Notes — Projectile Motion Subtopic

by Notetube Official•191 words•228 views
Cue / QuestionNotes
What stays constant in projectile motion?Horizontal velocity ux=ucos⁡θu_x = u\cos\theta — no horizontal force
What changes in projectile motion?Vertical velocity: vy=usin⁡θ−gtv_y = u\sin\theta - gt
At highest point, what is vertical velocity?vy=0v_y = 0; only ucos⁡θu\cos\theta remains
If θ = 45°, what is range?Rmax=u2gR_{max} = \frac{u^2}{g} (maximum possible range)
Derive time to reach highest pointvy=0⇒tup=usin⁡θg=T2v_y = 0 \Rightarrow t_{up} = \frac{u\sin\theta}{g} = \frac{T}{2}
What is the trajectory equation?y=xtan⁡θ−gx22u2cos⁡2θy = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta} (parabola)
For θ = 30° and θ = 60° same u, compare HH60=3×H30H_{60} = 3 \times H_{30}; ratio = tan^{2}\60°/tan^{2}\30° = 3
What is the angle for range = height?Solve: u2sin⁡2θg=u2sin⁡2θ2g\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 ucos⁡θu\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.

Like these notes? Save your own copy and start studying with NoteTube's AI tools.

Sign up free to clone these notes