Project Euler

Firecracker

A firecracker explodes at height h_0 = 100 m above flat ground, ejecting fragments in every direction at speed v_0 = 20 m/s. With gravity g = 9.81 m/s^2 and no air resistance, determine the volume...

Source sync Apr 19, 2026

Problem #0317

Level Level 08

Solved By 3,105

Languages C++, Python

Answer 1856532.8455

Length 260 words

geometryalgebramodular_arithmetic

A firecracker explodes at a height of $100\,\mathrm {m}$ above level ground. It breaks into a large number of very small fragments, which move in every direction; all of them have the same initial velocity of $20\,\mathrm {m/s}$.

We assume that the fragments move without air resistance, in a uniform gravitational field with $g=9.81\,\mathrm {m/s^2}$.

Find the volume (in $\metre ^3$) of the region through which the fragments move before reaching the ground. Give your answer rounded to four decimal places.

Problem 317: Firecracker

Mathematical Foundation

Lemma 1 (Parametric Trajectory). A fragment launched at polar angle $\alpha$ from the upward vertical traces the trajectory (in cylindrical coordinates):

r(t) = v_0 \sin\alpha \cdot t, \qquad z(t) = h_0 + v_0 \cos\alpha \cdot t - \tfrac{1}{2}g t^2.

Proof. The horizontal component of velocity is $v_0 \sin\alpha$ ; the vertical component is $v_0\cos\alpha$ (upward). Under constant gravitational acceleration $g$ downward, Newton’s second law gives $\ddot{z} = -g$ . Integrating with initial conditions $r(0) = 0$ , $z(0) = h_0$ , $\dot{r}(0) = v_0\sin\alpha$ , $\dot{z}(0) = v_0\cos\alpha$ yields the stated equations. $\square$

Theorem 1 (Envelope of Trajectories). The boundary of the reachable region is the paraboloid:

z_{\max}(r) = Z - \frac{g\,r^2}{2v_0^2}, \qquad Z := h_0 + \frac{v_0^2}{2g}.

Proof. Eliminate $t$ from the parametric equations. Set $u = \cot\alpha$ so that $\sin^2\alpha = 1/(1+u^2)$ :

z = h_0 + ru - \frac{gr^2}{2v_0^2}(1 + u^2).

This is quadratic in $u$ with negative leading coefficient $-gr^2/(2v_0^2)$ . The maximum over $u$ occurs at $\partial z/\partial u = 0$ :

r - \frac{gr^2}{v_0^2}\,u = 0 \implies u^* = \frac{v_0^2}{gr}.

Substituting $u^*$ :

\begin{align} z_{\max}(r) &= h_0 + r \cdot \frac{v_0^2}{gr} - \frac{gr^2}{2v_0^2}\!\left(1 + \frac{v_0^4}{g^2 r^2}\right) \\ &= h_0 + \frac{v_0^2}{g} - \frac{gr^2}{2v_0^2} - \frac{v_0^2}{2g} \\ &= h_0 + \frac{v_0^2}{2g} - \frac{gr^2}{2v_0^2} = Z - \frac{gr^2}{2v_0^2}. \quad \square \end{align}

Theorem 2 (Volume of the Reachable Region). The volume of the region is:

V = \frac{\pi v_0^2\,Z^2}{g}, \qquad Z = h_0 + \frac{v_0^2}{2g}.

Proof. The reachable region is bounded above by $z = z_{\max}(r)$ and below by $z = 0$ . By axial symmetry, apply the disc method. From the envelope, $r^2(z) = \frac{2v_0^2}{g}(Z - z)$ for $0 \le z \le Z$ :

\begin{align} V &= \pi \int_0^{Z} r^2(z)\,dz = \frac{2\pi v_0^2}{g}\int_0^{Z}(Z - z)\,dz \\ &= \frac{2\pi v_0^2}{g}\left[Zz - \frac{z^2}{2}\right]_0^Z = \frac{2\pi v_0^2}{g}\cdot\frac{Z^2}{2} = \frac{\pi v_0^2 Z^2}{g}. \quad \square \end{align}

Corollary (Numerical Evaluation).

Z = 100 + \frac{400}{2 \times 9.81} = 100 + \frac{200}{9.81} = 100 + 20.387359\ldots = 120.387359\ldots

V = \frac{\pi \times 400 \times (120.387359\ldots)^2}{9.81} = 1856532.8455\ldots

Proof. Direct substitution of $h_0 = 100$ , $v_0 = 20$ , $g = 9.81$ into Theorem 2. $\square$

Editorial

A firecracker at height h0 = 100m ejects fragments at v0 = 20 m/s in all directions. Find the volume of the reachable region (no air resistance, g = 9.81 m/s^2). The envelope of all parabolic trajectories forms a paraboloid of revolution: z_max(r) = Z - gr^2 / (2v0^2) where Z = h0 + v0^2 / (2*g) is the maximum height. Volume = pi * v0^2 * Z^2 / g. We return round(V, 4).

Pseudocode

Input: h0 = 100, v0 = 20, g = 9.81
Output: volume V to 4 decimal places
Z = h0 + v0^2 / (2 * g)
V = pi * v0^2 * Z^2 / g
Return round(V, 4)

Complexity Analysis

Time: $O(1)$ — direct closed-form evaluation.
Space: $O(1)$ .

Answer

\boxed{1856532.8455}

C++ project_euler/problem_317/solution.cpp

#include <bits/stdc++.h>
using namespace std;

/*
 * Problem 317: Firecracker
 *
 * V = pi * v0^2 * Z^2 / g
 * where Z = h0 + v0^2 / (2*g)
 *
 * h0 = 100, v0 = 20, g = 9.81
 */

int main(){
    double h0 = 100.0;
    double v0 = 20.0;
    double g = 9.81;

    double Z = h0 + v0 * v0 / (2.0 * g);
    double V = M_PI * v0 * v0 * Z * Z / g;

    printf("%.4f\n", V);
    return 0;
}

Python project_euler/problem_317/solution.py

"""
Problem 317: Firecracker

A firecracker at height h0 = 100m ejects fragments at v0 = 20 m/s in all directions.
Find the volume of the reachable region (no air resistance, g = 9.81 m/s^2).

The envelope of all parabolic trajectories forms a paraboloid of revolution:
    z_max(r) = Z - g*r^2 / (2*v0^2)
where Z = h0 + v0^2 / (2*g) is the maximum height.

Volume = pi * v0^2 * Z^2 / g
"""

import math

# Optional visualization