Physics 1 — Motion & Energy

Projectile motion

continues from lesson 2 — values defined earlier in the course stay live here

Launch at an angle and the motion splits cleanly in two: horizontal velocity never changes; vertical velocity is free fall. The two only share the clock.

v_{launch} = 20

launch speed (m/s)

\theta := 35 deg

= 35 deg launch angle

R_{range} = \frac{v_{launch}^{2} \cdot \mathrm{sin}\left(2 \cdot \theta\right)}{9.81}

38.3157

horizontal range (m)

h_{max} = \frac{\left(v_{launch} \cdot \mathrm{sin}\left(\theta\right)\right)^{2}}{2 \cdot 9.81}

6.7072

peak height (m)

x * tan(theta) - 9.81 * x^2 / (2 * (v_launch * cos(theta))^2)

the flight path y(x) — a parabola

Try theta := 45 deg — the textbook maximum-range angle — then 30° and 60°, which land in the same spot (sin(2θ) is symmetric about 45°).