Motion in Two or Three Dimensions - AP Physics C: Mechanics

$t_{\text{flight}} = \frac{2v_0 \sin(\theta)}{g}$. Time for projectile to return to launch height.

$\omega = \frac{2\pi}{T}$. Angular velocity is $2\pi$ radians per period $T$.

Toward the center of the circle. Centripetal acceleration always points toward the rotation center.

$\textbf{r} = x\textbf{i} + y\textbf{j}$. Position vector using unit vectors $\textbf{i}$ and $\textbf{j}$ for x and y components.

$F_c = 18 \text{ N}$. Using $F_c = \frac{mv^2}{r} = \frac{2 \times 9}{1} = 18$ N.

$R = 10.2 \text{ m}$. Range formula with $v_0 = 10$ m/s and $\theta = 45°$.

$t = 2 \text{ s}$. Using $h = \frac{1}{2}gt^2$: $t = \sqrt{\frac{2h}{g}} = 2$ s.

$v = 10 \text{ m/s}$. Pythagorean theorem: $\sqrt{6^2 + 8^2} = 10$ m/s.

$v = 5 \text{ m/s}$. Pythagorean theorem: $\sqrt{3^2 + 4^2} = 5$ m/s.

$v = \frac{2\text{π}r}{T}$. Circumference divided by period gives tangential velocity in circular motion.

$a_c = \frac{v^2}{r}$. Acceleration directed toward center equals velocity squared over radius.

$R = \frac{v_0^2 \text{sin}(2\theta)}{g}$. Maximum range occurs at 45° launch angle using trigonometric identity.

$g = 9.8 \text{ m/s}^2$. Standard value for Earth's gravitational acceleration near surface.

$y = v_{0y}t + \frac{1}{2}a_yt^2$. Kinematic equation for vertical motion with gravitational acceleration.

$x = v_{0x}t + \frac{1}{2}a_xt^2$. Kinematic equation for horizontal motion with constant acceleration.

$\textbf{a} = \frac{d\textbf{v}}{dt}$. Time derivative of velocity vector gives acceleration vector.

$\textbf{v} = \frac{d\textbf{r}}{dt}$. Time derivative of position vector gives velocity vector.

$\textbf{r} = x\textbf{i} + y\textbf{j} + z\textbf{k}$. Adds $\textbf{k}$ unit vector for the z-component in three dimensions.

$h = 11.5 \text{ m}$. Using $h = \frac{v_0^2}{2g}$ for maximum height in vertical motion.

$\theta = \omega t$. Angular displacement equals angular velocity times time.

$\textbf{v}_{\text{avg}} = \frac{\textbf{r}_2 - \textbf{r}_1}{t_2 - t_1}$. Change in position divided by change in time for vector quantities.

$R_{\text{max}} = 10.2 \text{ m}$. Maximum range occurs at 45° launch angle.

$x = 15 \text{ m}$. Horizontal displacement with constant velocity: $x = v_x \times t$.

$F_c = ma_c$. Newton's second law applied to centripetal motion.

$v = r\omega$. Linear velocity equals radius times angular velocity.

$\textbf{v}_{\text{rel}} = \textbf{v}_A - \textbf{v}_B$ . Velocity of object A relative to object B using vector subtraction.

$v_y = -19.6 \text{ m/s}$. Using $v_y = v_{0y} + a_yt = 0 + (-9.8)(2) = -19.6$ m/s.

$v = v_{0x}\textbf{i}$. At peak, vertical velocity is zero, only horizontal remains.

$\omega = \frac{\text{π}}{2} \text{ rad/s}$. Angular velocity equals $\frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2}$ rad/s.