Topic 1.5 Vectors and Motion in Two Dimensions: analyze two-dimensional motion by resolving into independent perpendicular components, apply this to projectile motion, and use vector calculus for general planar motion.

What this topic is asking

The College Board (Topic 1.5) wants you to analyze two-dimensional motion by resolving it into two independent one-dimensional problems on perpendicular axes, and to apply this to projectile motion, where the horizontal velocity is constant and the vertical motion is free fall. In AP Physics C the position, velocity and acceleration are vectors $\vec{r}(t)$, $\vec{v}(t)$, $\vec{a}(t)$ related by differentiation, and a projectile's parabolic path follows from eliminating time between the two component equations.

Two dimensions as two independent problems

This independence is the central idea. A projectile's horizontal motion does not "know" about its vertical motion: a bullet fired horizontally and a bullet dropped from the same height hit the ground at the same instant, because their vertical motions are identical. The strategy for any 2D problem is to resolve the initial velocity into components, write the kinematics for each axis separately, and use $t$ as the link.

The vector description

In AP Physics C the cleanest statement uses vectors. The position is $\vec{r}(t) = x(t)\hat{\imath} + y(t)\hat{\jmath}$, and the velocity and acceleration are its derivatives, taken component by component:

Because differentiation acts on each component separately, a 2D problem is genuinely two 1D problems. This formalism also handles non-projectile planar motion (a particle whose $x$ and $y$ are arbitrary functions of time), which AP Physics C does test: differentiate each component to get the velocity and acceleration vectors at any instant.

Projectile motion

A projectile moves under gravity alone after launch. Resolving the launch velocity into $v_{0x} = v_0\cos\theta$ and $v_{0y} = v_0\sin\theta$:

• Horizontal: $a_x = 0$, so $v_x = v_{0x}$ is constant and $x = v_{0x}\,t$.
• Vertical: $a_y = -g$, so $v_y = v_{0y} - gt$ and $y = v_{0y}t - \tfrac{1}{2}gt^2$.

From these come the standard results. At the top of the path $v_y = 0$, giving the maximum height $H = v_{0y}^2/(2g)$. The time of flight for a launch and landing at the same height is $t = 2v_{0y}/g$. The range is $R = v_{0x}\,t = v_0^2\sin(2\theta)/g$, which is greatest at $\theta = 45^\circ$. Eliminating $t$ between the two equations gives the trajectory $y = (\tan\theta)x - \dfrac{g}{2v_0^2\cos^2\theta}x^2$, a downward parabola.

Try this

Q1. A projectile is launched at $40$ m/s at $53^\circ$ above the horizontal ($g = 9.8$ m/s squared). Calculate the maximum height. [3 points]

• Cue. $v_{0y} = 40\sin 53^\circ = 32$ m/s; $H = v_{0y}^2/(2g) = 32^2/(2\times 9.8) = 52$ m.

Q2. State the launch angle that maximizes the range of a projectile over level ground, and explain why. [2 points]

• Cue. $45^\circ$: the range $R = v_0^2\sin(2\theta)/g$ is greatest when $\sin(2\theta) = 1$, i.e. $2\theta = 90^\circ$.