⊕Read Tipler and Mosca (2004) chapter 3 (Motion in 2 and 3D); Pelcovits and Farkas (2024) chapter 2 (Kinematics). See also https://www.youtube.com/watch?v=ZM8ECpBuQYE

The general idea is to split 2D or 3D motion up into components. Since we already know how to solve 1D motion problems, divide and conquer to solve 2D and 3D problems.

Projectile motion

Projectile motion can be thought of as a combination of constant velocity motion in the horizontal direction: \[\begin{aligned} x &= v_h t + x_0 \\ v_h &= \text{constant} \\ a_h &= 0 \end{aligned}\] and motion at constant acceleration in the vertical direction. \[\begin{aligned} y &= \dfrac{1}{2}a t^2 + v_{v,0} t + y_0 \\ v_v &= a_v t + v_{v,0}\\ a_v &= \text{constant} \end{aligned}\] where \(a_v=-g=\qty{-9.81}{\meter\per\second\squared}\) is the acceleration of gravity.

Recall that a velocity \(V\) at an incline \(\theta\) to the horizontal can be resolved into horizontal and vertical components \[\begin{aligned} V_h &= V \cos{\theta} \\ V_v &= V \sin{\theta} \end{aligned}\] For a projectile launched from the ground at initial position \((0,0)\): \[\begin{aligned} x &= V \cos{\theta} t \\ y &= -\dfrac{1}{2} g t^2 + V \sin{\theta} t. \end{aligned}\] The time in flight is \(t_f = 2 \dfrac{V \sin{\theta}}{g}\), so the range is \[x_f = \dfrac{V^2}{g} 2 \cos{\theta} \sin{\theta},\] or, using the trigonometric identity \(\sin{2\theta} = 2\sin{\theta}\cos{\theta}\): \[x_f = \dfrac{V^2}{g} \sin{2 \theta}.\]

The maximum range should be \(\dfrac{V^2}{g}\) at \(\theta=\ang{45}\). This is twice the maximum height attained when firing verticallyTartaglia’s Law, see https://en.wikipedia.org/wiki/Nicolo_Tartaglia

, \(h=\dfrac{V^2}{2g}\).

Launching from \(y_0\neq 0\)

For a projectile launched at a slight height \(y_0\) from the ground, the expression for range is slightly more complicated as we can no longer easily factor when finding the time in flight. Instead, we must use the quadratic formulaFor the quadratic \(0=At^2 + Bt + C\), the roots are \(t = \dfrac{-B\pm\sqrt{B^2-4AC}}{2A}\).

\[\begin{aligned} y &= \dfrac{1}{2}a t^2 + v_{v,0} t + y_0 \\ 0 &= \dfrac{1}{2}g t^2 + V \cos{\theta} t + y_0 \\ t_f &= \dfrac{-V \cos{\theta} \pm \sqrt{ (V \cos{\theta})^2 - 4 (\frac{1}{2}g) y_0}}{2 \frac{1}{2} g} \\ &= \dfrac{-V \cos{\theta} - \sqrt {V^2\cos^2{\theta} + - 2 g y_0}}{g} \\ x_f &= V \cos{\theta} t_f. \end{aligned}\] where \(V\) is the initial velocity, \(\theta\) is the angle, \(y_0\) is the initial height of the ball, and \(g=\qty{-9.81}{\meter\per\second\squared}\) is the acceleration of gravity. Note the maximum range no longer appears at \(\theta=\ang{45}\).

Similarly, when firing at a target not at \(y=0\), solutions must be adjusted. This must usually be done using numerical methods, in particular there is a “method of shooting,” an algorithm in which a particular angle \(\theta\) is assumed, the range found, and \(\theta\) adjusted mathematically until the correct endpoint is found.

Other 2D motion

The motion of billiard balls, knock hockey pucks, etc can be broken into \(x\) and \(y\) components and analyzed using 1D kinematics. At walls, we often reflect objects by reversing one of the components at the instant they bounce off the wall. Navigation of a ship at sea, a robot driving on a playing field, etc is often done as a 2D problem.

Typical problems

Typical testable problems might include things like ski jumpers, basketball shots, cricket or baseball hits, cannonballs, marbles, spitting fish, tossed dwarves...

Navigation links

Previous: Motion in 1D
Table of contents
Next: Relative motion

References

Pelcovits, Robert A, and Joshua Farkas. 2024. Barron’s AP Physics c Premium. Kaplan North America.

Tipler, Paul A, and Gene Mosca. 2004. Physics for Scientists and Engineers. 5th ed. W H Freeman; Company.