Research & Engineering Archive

DYN 3. Projectile Motion Edited

By Jingnan Huang · March 24, 2025 · 322 Words

Last Edit: 3/24/25

Assumptions
#

Axis	Initial Velocity	Acceleration
x	$v_0 \cos \theta$	0
y	$v_0 \sin \theta$	$\downarrow g, , -g \uparrow$

Axis	Velocity ($v = v_0 + at$)	Position ($s = s_0 + v_0t + \frac{1}{2}at^2$)	No “t” ($v^2 = v_0^2 + 2a\Delta s)$
x	$v_x = v_0 \cos \theta$	$x = x_0 + (v_0 \cos \theta)t$	N/A
y	$v_y = v_0 \sin \theta - gt$	$y = y_0 + (v_0 \sin \theta)t - \frac{1}{2}gt^2$	$v_y^2 = (v_0 \sin \theta)^2 - 2g\Delta y$

$$ t=\frac{v_y}{g} $$

$$ \Delta y = \frac{v_{y0}^2}{2g} $$

Snowmobile is going 15 𝑚/𝑠 at point A. Find: The horizontal distance it travels (𝑅) and the time (𝑡) in the air

Axis	Velocity (\(v = v_0 + at\))	Position (\(s = s_0 + v_0t + \frac{1}{2}at^2\))	No “t” (\(v^2 = v_0^2 + 2a\Delta s)\)
x	\(v_x = v_0 \cos \theta\)	\(x = x_0 + (v_0 \cos \theta)t\)	N/A
y	\(v_y = v_0 \sin \theta - gt\)	\(y = y_0 + (v_0 \sin \theta)t - \frac{1}{2}gt^2\)	\(v_y^2 = (v_0 \sin \theta)^2 - 2g\Delta y\)

Axis	Initial Velocity	Acceleration
x	\(v_0 \cos \theta\)	0
y	\(v_0 \sin \theta\)	\(\downarrow g, , -g \uparrow\)