---
title: "Free Fall and Projectile Motion"
book: "High School Physics"
subject: physics
language: en
chapter: 26
exercises: 15
source: https://one-course.com/books/physics/2/en/chapter/26-free-fall-and-projectile-motion
---

# Chapter 26 — Free Fall and Projectile Motion

Every jet of a garden hose draws the same curve in the air. So does a basketball arcing toward the hoop, a long jumper, a cannonball: any body abandoned to its [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight). This chapter earns that curve — a parabola — from [Newton’s second law](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#thm-g12-newtons-laws-second) in two lines of calculus, reads apex, flight time and [range](#def-g12-projectile-motion-range) off its equation, then lets the air back in to see what it spoils.

## 26.1 Free fall

**Definition 26.1 (Free fall).**

A body is in *free fall* when the only [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) acting on it is its [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) $\vect P = m\vect g$: no support, no string, air resistance negligible. A dropped stone or a ball that has left the hand is, to good accuracy, in free fall — “fall” includes rising.

**Theorem 26.2 (Galileo’s law of free fall).**

In [free fall](#def-g12-projectile-motion-freefall), all bodies share the same acceleration, whatever their mass:

$$
\vect a = \vect g, \qquad g = 9.81\,\mathrm{m}/\mathrm{s}^{2},
$$

directed straight down: the [field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-field) strength $9.81\,\mathrm{N}/\mathrm{kg}$ of [Chapter 4](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#ch-g10-universal-gravitation) in its other outfit, $\mathrm{N}/\mathrm{kg} = \mathrm{m}/\mathrm{s}^{2}$.

**Proof.** [Newton’s second law](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#thm-g12-newtons-laws-second) ([Chapter 25](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#ch-g12-newtons-laws)) with the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) as only [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force): $m\vect a = m\vect g$, and the mass cancels. Hammer and feather differ only through the air. ∎

**Proposition 26.3 (Vertical fall from rest).**

Dropped from rest, with the fallen depth $h$ counted downward from the release point, a body in [free fall](#def-g12-projectile-motion-freefall) obeys

$$
a = g, \qquad v = g t, \qquad h = \tfrac12\, g t^2 .
$$

**Proof.** $v(t)$ is the antiderivative of the constant $a = g$ vanishing at $t = 0$ ([Chapter 24](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#ch-g12-kinematics-2d)); $h(t)$ that of $v = gt$ vanishing at $t = 0$. ∎

**Example 26.4 (Orders of magnitude).**

After $1$, $2$, $3$ seconds of [free fall](#def-g12-projectile-motion-freefall): $v = 9.8$, $19.6$, $29.4\,\mathrm{m}/\mathrm{s}$ ($35$, $71$, $106\,\mathrm{km}/\mathrm{h}$) and $h = 4.9$, $19.6$, $44.1\,\mathrm{m}$. From a $30\,\mathrm{m}$ rooftop: $t = \sqrt{2h/g} \approx 2.5\,\mathrm{s}$, arriving at $24\,\mathrm{m}/\mathrm{s}$ $\approx 87\,\mathrm{km}/\mathrm{h}$. Times grow like $\sqrt h$ — and past a few seconds the air refuses to stay negligible (last section).

**Example 26.5 (Vertical throw).**

Thrown straight up at $v_0$, axis $Oy$ upward from the hand: the same two antiderivatives give $v = v_0 - gt$ and $y = v_0 t - \tfrac12 gt^2$. The apex, where $v = 0$, comes at $t = v_0/g$, height $h = v_0^2/2g$ — the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) answer $v_0 = \sqrt{2gh}$ of [Chapter 18](https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation#ch-g11-mechanical-energy) run backward; thrown *down*, $v_0$ flips sign. For $15\,\mathrm{m}/\mathrm{s}$: apex $11.5\,\mathrm{m}$ at $1.53\,\mathrm{s}$, back at $3.06\,\mathrm{s}$, at $15\,\mathrm{m}/\mathrm{s}$.

## 26.2 The projectile: one law, two antiderivatives

**Definition 26.6 (Projectile).**

A *projectile* is a body launched with a velocity $\vect v_0$ and then left in [free fall](#def-g12-projectile-motion-freefall). Axes at the launch point: $Ox$ horizontal in the vertical plane of $\vect v_0$, $Oy$ vertical upward; the *launch angle* $\alpha$ of $\vect v_0$ above the horizontal fixes its coordinates $(v_0\cos\alpha,\; v_0\sin\alpha)$.

**Theorem 26.7 (Equations of motion).**

A [projectile](#def-g12-projectile-motion-projectile) launched from the origin with speed $v_0$ and angle $\alpha$ has $\vect a = (0,\,-g)$ at every instant, hence

$$
\vect v(t) = \bigl(v_0\cos\alpha,\; v_0\sin\alpha - g t\bigr),
\qquad
\begin{cases}
x(t) = v_0\cos\alpha\; t,\\[2pt]
y(t) = v_0\sin\alpha\; t - \tfrac12\, g t^2 .
\end{cases}
$$

**Proof.** [Newton’s second law](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#thm-g12-newtons-laws-second) gives $\vect a = \vect g = (0,-g)$. Antiderivatives axis by axis: $\vect v(t)$ from $\vect a$ with initial value $(v_0\cos\alpha,\, v_0\sin\alpha)$, then position from $\vect v$, from $(0,0)$. ∎

**Proposition 26.8 (Independence of the two motions).**

The horizontal motion is uniform — $v_x$ never changes — and the vertical motion is exactly the vertical throw of the previous section: neither equation mentions the other’s coordinate. In particular, a ball fired horizontally and a ball dropped at the same instant from the same height stay level with each other — they land together.

**Proof.** In [Theorem 26.7](#thm-g12-projectile-motion-equations), $x(t)$ involves only $v_0\cos\alpha$, $y(t)$ only $v_0\sin\alpha$ and $g$; both balls have $v_y(0) = 0$, so both obey $h = \tfrac12 gt^2$. ∎

![One ball dropped, one fired horizontally at the same instant: at each time t_1, t_2, t_3 the same height — speed delays nothing.](https://one-course.com/images/onecourse/chapters/physics-2/g12-projectile-motion/fig-c13d7b0707ad.svg)

*One ball dropped, one fired horizontally at the same instant: at each time $t_1$, $t_2$, $t_3$ the same height — speed delays nothing.*

**Remark 26.9 (The falling monkey).**

A zookeeper aims a dart *straight at* a monkey, who lets go of its branch at the shot. Bad move: dart and monkey each drop $\tfrac12 gt^2$ below their no-gravity positions — the aiming line, the branch — so they meet exactly, whatever the dart’s speed ([Exercise 26.15](#exo-g12-projectile-motion-15)).

## 26.3 The parabola, read in full

**Theorem 26.10 (Trajectory).**

Eliminating $t$ between $x(t)$ and $y(t)$: the [trajectory](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#def-g10-relative-motion-trajectory) of a [projectile](#def-g12-projectile-motion-projectile) ($0 \leq \alpha < 90{}^{\circ}$) is the downward-opening parabola

$$
y = x\tan\alpha \;-\; \frac{g}{2 v_0^2 \cos^2\alpha}\, x^2 .
$$

**Proof.** $t = x/(v_0\cos\alpha)$, substituted into $y(t)$ of [Theorem 26.7](#thm-g12-projectile-motion-equations). ∎

![The velocity along the flight: v_x never changes; v_y shrinks, vanishes at the apex, reverses — the arrow tilts, the parabola follows.](https://one-course.com/images/onecourse/chapters/physics-2/g12-projectile-motion/fig-9bca4402f8e5.svg)

*The velocity along the flight: $\vect v_x$ never changes; $\vect v_y$ shrinks, vanishes at the apex, reverses — the arrow tilts, the parabola follows.*

**Definition 26.11 (Range).**

For a [projectile](#def-g12-projectile-motion-projectile) over level ground, the *range* $R$ is the horizontal distance covered when it returns to its launch height; the *time of flight* $t_f$ is that trip’s duration.

**Proposition 26.12 (Apex, flight time, range).**

Over level ground, with $0 < \alpha < 90{}^{\circ}$:

$$
t_f = \frac{2 v_0 \sin\alpha}{g}, \qquad
h_{\text{apex}} = \frac{(v_0\sin\alpha)^2}{2g}
\ \ \text{reached at } t_f/2, \qquad
R = \frac{v_0^2 \sin(2\alpha)}{g}.
$$

The [range](#def-g12-projectile-motion-range) is greatest at $\alpha = 45{}^{\circ}$, where $R = v_0^2/g$; complementary angles $\alpha$ and $90{}^{\circ} - \alpha$ give the same [range](#def-g12-projectile-motion-range).

**Proof.** $y = 0$ gives $t\,(v_0\sin\alpha - \tfrac12 gt) = 0$: launch, and $t_f$. The apex, $v_y = 0$, sits at $t = v_0\sin\alpha/g = t_f/2$; substitute into $y(t)$. Then $R = x(t_f) = 2v_0^2\sin\alpha\cos\alpha/g =
v_0^2\sin(2\alpha)/g$. The sine peaks at $2\alpha = 90{}^{\circ}$, and $\sin\bigl(2(90{}^{\circ}-\alpha)\bigr) = \sin(180{}^{\circ} - 2\alpha) =
\sin(2\alpha)$. ∎

![Same launch speed v_0 = 10\, m/ s, five angles: 45 throws farthest; the pairs (15,75) and (30,60) land together — one flat and fast, one high and slow.](https://one-course.com/images/onecourse/chapters/physics-2/g12-projectile-motion/fig-4bb814d17b83.svg)

*Same launch speed $v_0 = 10\,\mathrm{m}/\mathrm{s}$, five angles: $45{}^{\circ}$ throws farthest; the pairs ($15{}^{\circ}$,$75{}^{\circ}$) and ($30{}^{\circ}$,$60{}^{\circ}$) land together — one flat and fast, one high and slow.*

**Example 26.13 (Goal kick).**

A goalkeeper strikes at $v_0 = 25\,\mathrm{m}/\mathrm{s}$, $\alpha = 45{}^{\circ}$: $R = 25^2/9.81 \approx 64\,\mathrm{m}$; real clearances land nearer $55\,\mathrm{m}$ — the air’s cut.

**Method 26.14 (Projectile bookkeeping).**

1. Axes at the launch point, $Ox$ horizontal, $Oy$ up; coordinates of $\vect v_0$ .
2. Check [free fall](#def-g12-projectile-motion-freefall) ( [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) only): then $\vect a = (0,-g)$ ; antiderivatives twice, initial conditions each time.
3. Question about a *place* : eliminate $t$ ( [trajectory](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#def-g10-relative-motion-trajectory) ). About an *instant* : keep $t$ .
4. Dictionary: apex $\Leftrightarrow v_y = 0$ ; launch height $\Leftrightarrow y = 0$ ; wall at $d \Leftrightarrow x = d$ ; floor $\ell$ below $\Leftrightarrow y = -\ell$ .
5. Sanity: flight symmetric about the apex; $R \leq v_0^2/g$ ; air only ever shortens.

## 26.4 What the air changes

**Definition 26.15 (Terminal speed).**

Air [resistance](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-resistance) opposes the velocity and grows with speed. A falling body therefore accelerates only until drag balances [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight); from there $\vect a = \vect 0$: it falls at a constant *terminal speed*.

**Example 26.16 (Terminal orders of magnitude).**

A flat-out skydiver: about $55\,\mathrm{m}/\mathrm{s}$ ($200\,\mathrm{km}/\mathrm{h}$). A raindrop: about $9\,\mathrm{m}/\mathrm{s}$ — in vacuum, $2000\,\mathrm{m}$ of fall would deliver it at $\sqrt{2gh} \approx 2.0 \times 10^{2}\,\mathrm{m}/\mathrm{s}$. A feather: about $1\,\mathrm{m}/\mathrm{s}$, reached within centimeters — the secret of its slowness: it spends its whole fall *at* [terminal speed](#def-g12-projectile-motion-terminal); the hammer never gets near its own.

**Remark 26.17 (Real ballistics).**

Drag pushes against the velocity all along the flight: the [range](#def-g12-projectile-motion-range) shortens, the apex lowers, and the symmetry breaks — the descent steeper than the climb, long shots visibly short of the parabola. The honest, quantitative treatment is a differential equation solved in the Year 1 volume; this year, the parabola is the right model for dense, moderately fast [projectiles](#def-g12-projectile-motion-projectile) — a basketball at $10\,\mathrm{m}/\mathrm{s}$, not a golf ball at $70\,\mathrm{m}/\mathrm{s}$.

![The same launch, with and without air: drag trims range and apex and lands the projectile steeper than it took off — the parabola’s symmetry is the first casualty.](https://one-course.com/images/onecourse/chapters/physics-2/g12-projectile-motion/fig-29c16ed662c2.svg)

*The same launch, with and without air: drag trims [range](#def-g12-projectile-motion-range) and apex and lands the [projectile](#def-g12-projectile-motion-projectile) steeper than it took off — the parabola’s symmetry is the first casualty.*

## 26.5 Exercises

**Exercise 26.1 ★.**

A stone dropped from a bridge hits the water $2.0\,\mathrm{s}$ later: height of the bridge, and impact speed in $\mathrm{m}/\mathrm{s}$ and $\mathrm{km}/\mathrm{h}$?

**Solution of Exercise 26.1.**

$h = \tfrac12 \times 9.81 \times 2.0^2 = 19.6\,\mathrm{m}$; $v = gt = 19.6\,\mathrm{m}/\mathrm{s} \approx 71\,\mathrm{km}/\mathrm{h}$.

**Exercise 26.2 ★.**

On the Moon, an astronaut dropped a hammer and a feather together: they landed together. Why — and why does the feather lose so badly on Earth?

**Solution of Exercise 26.2.**

No air on the Moon: both in [free fall](#def-g12-projectile-motion-freefall), $m\vect a = m\vect g$ gives $\vect a = \vect g$ for any mass. On Earth, drag on the feather is comparable to its tiny [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight): it rides at a [terminal speed](#def-g12-projectile-motion-terminal) of about $1\,\mathrm{m}/\mathrm{s}$ while the hammer barely notices the air.

**Exercise 26.3 ★.**

Neglecting air, compute fall time and arrival speed from $1.0\,\mathrm{m}$, $10\,\mathrm{m}$, $100\,\mathrm{m}$. How do the times grow with the height?

**Solution of Exercise 26.3.**

$t = \sqrt{2h/g} = 0.45$, $1.4$, $4.5\,\mathrm{s}$; $v = gt = 4.4$, $14$, $44\,\mathrm{m}/\mathrm{s}$. Times grow like $\sqrt h$: $\times 100$ in height is only $\times 10$ in time.

**Exercise 26.4 ★.**

A ball is thrown straight up at $15\,\mathrm{m}/\mathrm{s}$: apex height and time, total time back to the hand, and return speed?

**Solution of Exercise 26.4.**

$h = v_0^2/2g = 11.5\,\mathrm{m}$ at $t = v_0/g = 1.53\,\mathrm{s}$; back at $3.06\,\mathrm{s}$, at $15\,\mathrm{m}/\mathrm{s}$ again (symmetry).

**Exercise 26.5 ★.**

A marble rolls off a table of height $0.80\,\mathrm{m}$ at $2.5\,\mathrm{m}/\mathrm{s}$. Find the fall time and the landing distance from the edge. Does the fall time depend on the marble’s speed?

**Solution of Exercise 26.5.**

$t = \sqrt{2 \times 0.80/9.81} = 0.40\,\mathrm{s}$; $x = 2.5 \times 0.40 = 1.0\,\mathrm{m}$. No: the fall time belongs to the vertical motion alone (independence).

**Exercise 26.6 ★★.**

From the five-angle figure of the course: which pairs of [launch angles](#def-g12-projectile-motion-projectile) share their [range](#def-g12-projectile-motion-range), and which angle wins? For a [range](#def-g12-projectile-motion-range) below the maximum, which of the two possible angles flies longer? Justify with $t_f$.

**Solution of Exercise 26.6.**

$(15{}^{\circ},75{}^{\circ})$ and $(30{}^{\circ},60{}^{\circ})$ share their [range](#def-g12-projectile-motion-range); $45{}^{\circ}$ wins. The steeper of a complementary pair flies longer: $t_f = 2v_0\sin\alpha/g$ grows with $\alpha$.

**Exercise 26.7 ★★.**

From a $20\,\mathrm{m}$ balcony a ball is thrown straight *down* at $5.0\,\mathrm{m}/\mathrm{s}$. Find the time to the ground and the impact speed; check the latter with an [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) balance ([Chapter 18](https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation#ch-g11-mechanical-energy)).

**Solution of Exercise 26.7.**

$4.905\,t^2 + 5.0\,t - 20 = 0$ gives $t = (-5 + \sqrt{25 + 392.4})/9.81 = 1.57\,\mathrm{s}$; $v = 5.0 + 9.81 \times 1.57 = 20.4\,\mathrm{m}/\mathrm{s}$. [Energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy): $v = \sqrt{5.0^2 + 2 \times 9.81 \times 20} = 20.4\,\mathrm{m}/\mathrm{s}$ — same.

**Exercise 26.8 ★★.**

A [projectile](#def-g12-projectile-motion-projectile) leaves level ground at $40\,\mathrm{m}/\mathrm{s}$, $\alpha = 30{}^{\circ}$: [time of flight](#def-g12-projectile-motion-range), apex height and [range](#def-g12-projectile-motion-range)?

**Solution of Exercise 26.8.**

$t_f = 2 \times 40 \times \sin30{}^{\circ}/9.81 = 4.1\,\mathrm{s}$; $h = (40\sin30{}^{\circ})^2/(2 \times 9.81) = 20.4\,\mathrm{m}$; $R = 40^2 \sin60{}^{\circ}/9.81 = 1.4 \times 10^{2}\,\mathrm{m}$.

**Exercise 26.9 ★★.**

Model a long jumper as a [projectile](#def-g12-projectile-motion-projectile): takeoff at $10.5\,\mathrm{m}/\mathrm{s}$, $\alpha = 20{}^{\circ}$. Compute the [range](#def-g12-projectile-motion-range); champions jump nearly $9\,\mathrm{m}$ — name two ingredients the point model leaves out.

**Solution of Exercise 26.9.**

$R = 10.5^2\sin40{}^{\circ}/9.81 \approx 7.2\,\mathrm{m}$. Left out: the [center of mass](https://one-course.com/books/physics/2/en/chapter/16-forces-and-motion#def-g11-forces-and-motion-com) takes off high (extended body) and lands low and far forward (legs thrown ahead), and the jumper is no point — the missing $1.7\,\mathrm{m}$ live in that geometry, not in the parabola.

**Exercise 26.10 ★★.**

At $1.2\,\mathrm{m}$ above flat ground, one bullet is dropped and an identical one fired horizontally at $300\,\mathrm{m}/\mathrm{s}$. Which lands first? Compute the fall time and the fired bullet’s landing distance. What principle is tested?

**Solution of Exercise 26.10.**

Together: both have $v_y(0) = 0$, so the same $t = \sqrt{2 \times 1.2/9.81} = 0.49\,\mathrm{s}$; the fired one lands $300 \times 0.49 \approx 1.5 \times 10^{2}\,\mathrm{m}$ away. It tests the independence of horizontal and vertical motions.

**Exercise 26.11 ★★.**

Show algebraically that $\alpha$ and $90{}^{\circ} - \alpha$ share their [range](#def-g12-projectile-motion-range), then find the two angles landing a $25\,\mathrm{m}/\mathrm{s}$ [projectile](#def-g12-projectile-motion-projectile) $50\,\mathrm{m}$ away.

**Solution of Exercise 26.11.**

$\sin\bigl(2(90{}^{\circ}-\alpha)\bigr) = \sin(180{}^{\circ}-2\alpha) =
\sin(2\alpha)$: same $R$. Here $\sin(2\alpha) = gR/v_0^2 =
9.81 \times 50/625 = 0.785$: $\alpha = 25.9{}^{\circ}$ or $64.1{}^{\circ}$.

**Exercise 26.12 ★★★.**

Water leaves a garden hose at $8.0\,\mathrm{m}/\mathrm{s}$: greatest horizontal reach? greatest vertical height? the two nozzle angles watering a flowerbed $5.0\,\mathrm{m}$ away?

**Solution of Exercise 26.12.**

$R_{\max} = v_0^2/g = 6.5\,\mathrm{m}$ (at $45{}^{\circ}$); $h_{\max} = v_0^2/2g = 3.3\,\mathrm{m}$ (straight up). $\sin(2\alpha) = 9.81 \times 5.0/64 = 0.766$: $\alpha = 25.0{}^{\circ}$ or $65.0{}^{\circ}$.

**Exercise 26.13 ★★★.**

A ball is kicked from the ground at $18\,\mathrm{m}/\mathrm{s}$, $\alpha = 35{}^{\circ}$, toward a $4.0\,\mathrm{m}$ hedge $15\,\mathrm{m}$ away. Using the [trajectory](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#def-g10-relative-motion-trajectory) equation, decide whether it clears, and by how much.

**Solution of Exercise 26.13.**

$y(15) = 15\tan35{}^{\circ} - \dfrac{9.81 \times 15^2}
{2 \times 18^2 \cos^235{}^{\circ}} = 10.5 - 5.1 = 5.4\,\mathrm{m}$: it clears the $4.0\,\mathrm{m}$ hedge by about $1.4\,\mathrm{m}$.

**Exercise 26.14 ★★★.**

An $80\,\mathrm{kg}$ skydiver falls flat at [terminal speed](#def-g12-projectile-motion-terminal) $55\,\mathrm{m}/\mathrm{s}$. (a) The drag [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force)? (b) Sketch the shape of $v(t)$ from the jump. (c) A raindrop falls from $2.0\,\mathrm{km}$: vacuum arrival speed vs. its real $9\,\mathrm{m}/\mathrm{s}$?

**Solution of Exercise 26.14.**

(a) Drag balances [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight): $80 \times 9.81 \approx 7.8 \times 10^{2}\,\mathrm{N}$. (b) $v(t)$ rises with ever-smaller slope and levels off at $55\,\mathrm{m}/\mathrm{s}$. (c) Vacuum: $\sqrt{2 \times 9.81 \times 2000} \approx 2.0 \times 10^{2}\,\mathrm{m}/\mathrm{s}$, twenty times the real $9\,\mathrm{m}/\mathrm{s}$ — raindrops spend nearly the whole fall at [terminal speed](#def-g12-projectile-motion-terminal).

**Exercise 26.15 ★★★.**

The falling monkey, proved: a dart leaves the origin aimed exactly at a monkey at distance $D$, height $H$ ($\tan\alpha = H/D$), who drops at $t = 0$. Show that when the dart reaches $x = D$ its height equals the monkey’s — whatever $v_0$.

**Solution of Exercise 26.15.**

At $t^* = D/(v_0\cos\alpha)$ the dart is at height $v_0\sin\alpha\, t^* - \tfrac12 g t^{*2} = D\tan\alpha - \tfrac12 g
t^{*2} = H - \tfrac12 g t^{*2}$, exactly the dropped monkey’s $y = H - \tfrac12 g t^{*2}$. Both hang $\tfrac12 gt^2$ below their no-gravity spots — they meet for any $v_0$.

## 26.6 Problem: The Basketball Buzzer-Beater

**Problem 26.1.**

Weekend problem — the basketball buzzer-beater: a free throw dissected arrow by arrow, fingertips cleared by centimeters, a full-court heave weighed against what an arm can give, a lob over the backboard — and a verdict on the most theatrical shot in sport

The rim of a basketball hoop is a horizontal ring $3.05\,\mathrm{m}$ above the floor; Nora releases every shot at $2.00\,\mathrm{m}$, so with axes at the release point ($Ox$ horizontal toward the hoop, $Oy$ up) the rim center sits $1.05\,\mathrm{m}$ above the origin. Air is neglected except where stated.

**Part I — The free throw.** Nora shoots from $d = 4.20\,\mathrm{m}$ (horizontal) with [launch angle](#def-g12-projectile-motion-projectile) $\alpha = 50{}^{\circ}$.

1. What [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) act once released? Give $\vect a$ , citing the law used.
2. Derive the coordinates of $\vect v(t)$ .
3. Derive $x(t)$ and $y(t)$ .
4. Eliminate $t$ to obtain the [trajectory](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#def-g10-relative-motion-trajectory) $y(x)$ .
5. The ball must pass the rim center, $y(d) = 1.05\,\mathrm{m}$ : solve for $v_0$ , in $\mathrm{m}/\mathrm{s}$ and $\mathrm{km}/\mathrm{h}$ .

**Part II — Over the fingertips, into the ring.**

6. How long does the ball take to reach the rim?
7. Find the apex time and height; check the ball is already descending at the rim.
8. A defender $0.90\,\mathrm{m}$ in front of Nora stretches fingertips to $2.60\,\mathrm{m}$ . By how much does the ball clear them?
9. Compute $v_x$ and $v_y$ at the rim, the speed there, and the angle of the velocity below the horizontal.
10. Why must a ball arrive *descending* — and why steep?

**Part III — The full-court heave.** Buzzer about to [sound](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-sound), Nora is $26.0\,\mathrm{m}$ from the far rim.

11. The $1.05\,\mathrm{m}$ rise is small against $26.0\,\mathrm{m}$ : justify a level-ground model, and derive the [time of flight](#def-g12-projectile-motion-range) from $y(t_f) = 0$ .
12. Deduce $R = v_0^2\sin(2\alpha)/g$ and the angle of greatest [range](#def-g12-projectile-motion-range) .
13. Compute the $v_0$ needed to cover $26.0\,\mathrm{m}$ at $\alpha = 45{}^{\circ}$ .
14. A strong player can hurl a basketball at about $17.5\,\mathrm{m}/\mathrm{s}$ . What is her maximum [range](#def-g12-projectile-motion-range) — is the shot humanly feasible?
15. At $17.5\,\mathrm{m}/\mathrm{s}$ , find the *two* angles landing the ball $26.0\,\mathrm{m}$ away, and their flight times: with $2.0\,\mathrm{s}$ on the clock, which one beats the buzzer?

**Part IV — The lob over the backboard.** From behind the baseline Nora lobs: release $6.0\,\mathrm{m}$ from the rim center, $\alpha = 75{}^{\circ}$; the board’s plane crosses her shot $0.45\,\mathrm{m}$ before the rim center, top edge $3.95\,\mathrm{m}$ above the floor.

16. Find the $v_0$ that drops the ball through the rim center.
17. Compute the ball’s height at the board’s plane ( $x = 5.55\,\mathrm{m}$ ): does it clear the top edge, by how much?
18. Find the apex height and flight time — why do crowds love it?
19. Qualitatively, what does air resistance change on these three shots, and which one does it threaten most?
20. *The buzzer-beater verdict* , two sentences: compare the speed the full-court shot demands with what an arm supplies, give the margin, and rule — possible or myth?

**Solution of Problem 26.1.**

**1.** [Weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) only (air neglected); [Newton’s second law](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#thm-g12-newtons-laws-second): $m\vect a = m\vect g$, so $\vect a = (0,\,-g)$.

**2.** $v_x = v_0\cos\alpha$, $v_y = v_0\sin\alpha - gt$.

**3.** $x = v_0\cos\alpha\, t$, $y = v_0\sin\alpha\, t - \tfrac12 gt^2$.

**4.** $t = x/(v_0\cos\alpha)$: $y = x\tan\alpha - g x^2/(2v_0^2\cos^2\alpha)$.

**5.** $v_0^2 = \dfrac{g d^2}{2\cos^2\alpha\,(d\tan\alpha -
1.05)} = \dfrac{173.0}{0.826 \times 3.96} = 52.9$: $v_0 = 7.28\,\mathrm{m}/\mathrm{s} \approx 26\,\mathrm{km}/\mathrm{h}$.

**6.** $t = d/(v_0\cos\alpha) = 4.20/4.68 = 0.90\,\mathrm{s}$.

**7.** Apex at $t = v_0\sin\alpha/g = 0.57\,\mathrm{s}$, height $2.00 + (v_0\sin\alpha)^2/2g = 2.00 + 1.58 = 3.58\,\mathrm{m}$; $0.57 < 0.90$: descending at the rim.

**8.** $y(0.90) = 0.90\tan50{}^{\circ} - 0.182 = 0.89\,\mathrm{m}$ above release, i.e. $2.89\,\mathrm{m}$ above the floor: $0.29\,\mathrm{m}$ above the fingertips.

**9.** $v_x = 4.68\,\mathrm{m}/\mathrm{s}$; $v_y = 5.57 - 9.81 \times 0.90 = -3.24\,\mathrm{m}/\mathrm{s}$; $v = 5.69\,\mathrm{m}/\mathrm{s}$, at $\arctan(3.24/4.68) = 34.7{}^{\circ}$ below the horizontal.

**10.** The rim is a horizontal ring: the ball must cross its plane from above. The steeper the entry, the larger the ring’s opening looks along the velocity — more margin for error.

**11.** $1.05/26.0 = 4\%$: negligible. $y(t_f) = 0$ gives $t_f\,(v_0\sin\alpha - \tfrac12 g t_f) = 0$, so $t_f = 2v_0\sin\alpha/g$.

**12.** $R = v_0\cos\alpha\, t_f = v_0^2\sin(2\alpha)/g$; maximal at $\alpha = 45{}^{\circ}$.

**13.** $v_0 = \sqrt{gR} = \sqrt{9.81 \times 26.0} =
16.0\,\mathrm{m}/\mathrm{s} \approx 57\,\mathrm{km}/\mathrm{h}$.

**14.** $R_{\max} = 17.5^2/9.81 = 31.2\,\mathrm{m} >
26.0\,\mathrm{m}$: feasible, with $5\,\mathrm{m}$ of [range](#def-g12-projectile-motion-range) to spare.

**15.** $\sin(2\alpha) = 255.1/306.3 = 0.833$: $\alpha = 28.2{}^{\circ}$ or $61.8{}^{\circ}$; flight times $2v_0\sin\alpha/g = 1.69\,\mathrm{s}$ and $3.14\,\mathrm{s}$. Only the flat shot beats the $2.0\,\mathrm{s}$ buzzer.

**16.** $v_0^2 = \dfrac{9.81 \times 36}{2\cos^275{}^{\circ}\,
(6.0\tan75{}^{\circ} - 1.05)} = \dfrac{353.2}{2.86} = 123.5$: $v_0 = 11.1\,\mathrm{m}/\mathrm{s}$.

**17.** $y(5.55) = 5.55\tan75{}^{\circ} - 18.26 = 2.45\,\mathrm{m}$ above release, i.e. $4.45\,\mathrm{m}$ above the floor: $0.50\,\mathrm{m}$ above the board’s top edge.

**18.** Apex $2.00 + (11.1\sin75{}^{\circ})^2/2g \approx
7.9\,\mathrm{m}$; flight $6.0/(11.1\cos75{}^{\circ}) = 2.1\,\mathrm{s}$ — two full seconds with the ball hanging $8\,\mathrm{m}$ up.

**19.** Drag shortens every shot and steepens the descent; the effect grows with speed, so the $17.5\,\mathrm{m}/\mathrm{s}$ full-court heave suffers most — its real [range](#def-g12-projectile-motion-range) falls a few meters short of $31\,\mathrm{m}$, eating into the margin.

**20.** *The buzzer-beater verdict*: the $26.0\,\mathrm{m}$ shot demands $16.0\,\mathrm{m}/\mathrm{s}$ and an arm supplies about $17.5\,\mathrm{m}/\mathrm{s}$, a [range](#def-g12-projectile-motion-range) margin of $5\,\mathrm{m}$ ($\approx 20\%$) that air resistance trims but does not erase. Possible — which is exactly why, a few times a season, it goes in.