Why a 30° Throw and a 60° Throw Land at the Same Spot

Both move in a plane with changing velocity. A thrown ball has constant acceleration $g$ downward and traces a parabola. A fan blade's tip moves at steady speed, yet its direction turns all the time, so it too is accelerating.

These are the two most important kinds of motion in a plane: projectile motion and uniform circular motion.

This part covers projectile equations, maximum range, centripetal acceleration, and the links between speed, angular speed, period and frequency.

**Splitting the launch velocity into $u\cos \theta$ horizontally (constant) and $u\sin \theta$ vertically (slowed by $g$) gives a parabolic path with $T=\frac{2u\sin \theta }{g}$, $H=\frac{u^2{\sin }^2\theta }{2g}$ and $R=\frac{u^2\sin 2\theta }{g}$.

Derivation.**

$$x=(u\cos \theta )ty=(u\sin \theta )t-\frac{1}{2}g{t}^2$$

Eliminating $t=\frac{x}{u\cos \theta }$ gives the trajectory:

$$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$$

Setting $y=0$ gives $T$; setting $v_y=0$ gives $H$; and $R=(u\cos \theta )T$.

Worked example. $u=20$ m/s at $30°$, with $g=10$ m/s${}^2$: $u_x\approx 17.32$ m/s, $u_y=10$ m/s.

$$T=\frac{2(10)}{10}=2\text{ s}H=\frac{10^2}{2(10)}=5\text{ m}R=\frac{400\sin 60°}{10}\approx 34.64\text{ m}$$

The trajectory is $y\approx 0.577x-\frac{x^2}{60}$, which indeed gives $y=0$ at $x\approx 34.64$ m.

An everyday example. Water from a garden hose held at an angle follows this parabola.

The substance. **At the highest point the velocity is $u\cos \theta$, not zero** — only the vertical part vanishes.

**Since $R=\frac{u^2\sin 2\theta }{g}$ is largest when $\sin 2\theta =1$, the maximum range $\frac{u^2}{g}$ occurs at $\theta =45°$, and complementary angles $\theta$ and $90°-\theta$ give equal ranges.

Worked example 1 — maximum range.** For $u=20$ m/s: $R_{max}=\frac{400}{10}=40$ m.

Worked example 2 — complementary angles.

- **At $30°$**: $R\approx 34.64$ m, $H=5$ m, $T=2$ s
- **At $60°$**: $R\approx 34.64$ m, $H=\frac{400(0.75)}{20}=15$ m, $T\approx 3.46$ s

Same range, but the steeper throw goes three times higher and stays up longer.

Worked example 3 — finding the angle. For a $20$ m range at $u=20$ m/s:

$$\sin 2\theta =\frac{20\times 10}{400}=0.5\Rightarrow 2\theta =30°\text{ or }150°\Rightarrow \theta =15°\text{ or }75°$$

Worked example 4 — minimum speed. To land a ball $40$ m away, $\frac{u^2}{g}\ge 40$, so $u\ge 20$ m/s.

An everyday example. In gully cricket, a flat hit and a high lofted hit can land the ball at the same spot, but the lofted one gives fielders more time to catch it.

The substance. **$45°$ is best only for level ground and negligible air resistance**; the range formula does not apply to a launch from a height.

**Angular velocity $\omega =\frac{d\theta }{dt}$ is the rate at which the radius sweeps angle, and a body moving in a circle at constant speed has acceleration $a_c=\frac{v^2}{r}={\omega }^{2}r$ directed towards the centre.

Linking linear and angular.** Arc length $s=r\theta$, so $v=r\omega$.

Derivation. Velocities ${\vec{v}}_1$ and ${\vec{v}}_2$ at two nearby points have the same magnitude $v$ but differ in direction by $\Delta \theta$. For small $\Delta \theta$:

$$|\Delta \vec{v}|=v\Delta \theta a=\underset{\Delta t\to 0}{\lim }\frac{v\Delta \theta }{\Delta t}=v\omega =\frac{v^2}{r}$$

As $\Delta t\to 0$, $\Delta \vec{v}$ points towards the centre.

Worked example. A car goes round a circular track of radius $50$ m at a steady $10$ m/s.

$$\omega =\frac{v}{r}=\frac{10}{50}=0.2\text{ rad/s}{a}_c=\frac{10^2}{50}=2{\text{ m/s}}^2$$

If the speed doubles to $20$ m/s, $a_c$ becomes $8$ m/s${}^2$ — four times larger.

An everyday example. A stone whirled on a string needs the string's constant inward pull; let go and it flies off along the tangent.

The substance. Uniform circular motion is accelerated motion, and because the direction of $\vec{a}$ keeps changing, the kinematic equations for constant acceleration do not apply.

**One revolution sweeps $2\pi$ radians in time period $T$, so $\omega =\frac{2\pi }{T}=2\pi f$ with frequency $f=\frac{1}{T}$, and the linear speed is $v=\omega r=\frac{2\pi r}{T}$.**

It also follows that $a_c={\omega }^{2}r=4{\pi }^2{f}^{2}r$.

Worked example 1 — a ceiling fan. Suppose a fan turns at $5$ revolutions per second with blade tips $0.6$ m from the centre.

$$\omega =2\pi (5)\approx 31.4\text{ rad/s}v=31.4\times 0.6\approx 18.8\text{ m/s}$$

$$a_c={\omega }^{2}r\approx 986.96\times 0.6\approx 592{\text{ m/s}}^{2}T=\frac{1}{5}=0.2\text{ s}$$

Worked example 2 — the rotating Earth. $T=24$ h $=86400$ s and the equatorial radius is about $6.4\times 10^6$ m.

$$\omega =\frac{2\pi }{86400}\approx 7.27\times 10^{-5}\text{ rad/s}v\approx 465\text{ m/s}{a}_c\approx 0.034{\text{ m/s}}^2$$

An everyday example. On a giant wheel at a mela, riders in outer seats and a child near the hub complete a turn in the same time, but the outer seats move much faster.

The substance. **Every point of a rotating rigid body has the same $\omega$**, while $v$ grows in proportion to distance from the axis.