{"id":1095,"date":"2018-03-09T07:15:01","date_gmt":"2018-03-09T14:15:01","guid":{"rendered":"http:\/\/blog.inkofpark.com\/?p=1095"},"modified":"2025-10-21T19:07:23","modified_gmt":"2025-10-22T01:07:23","slug":"launch-o-rocket","status":"publish","type":"post","link":"https:\/\/www.inkofpark.com\/?p=1095","title":{"rendered":"Launch-o-Rocket"},"content":{"rendered":"

School, from our house as the crow flies, is 5.73 km. If we neglect air resistance and deal strictly with ballistic flight then we can materialize a wonderful fantasy. Starting in the backyard, extending over the top of the house, is a launch-o-rocket<\/em>, a rail-like launcher that accelerates the school-bound student until he or she can cruise over the city and arrive without bother of traffic. Our charter is to find the acceleration of the student from the launch-o-rocket.<\/p>\n

\"launch-o-rocket\"<\/a><\/p>\n

Finding the Initial Velocity<\/h2>\n

We rely on the well-known fact that the maximum distance in a throw occurs when the departure angle is 45\u00b0. The vertical speed and the horizontal speed are equal. We denote these two identical speeds as $s$. Since distance is time multiplied by speed, the distance from home to school $d$ is
\n$$\u00a0 d = t\\cdot s.$$<\/p>\n

We know the distance $d = $ 5.73 km.<\/p>\n

Turning to the vertical speed, the student departs the launch-o-rocket with vertical speed $s$, but is immediately subject to gravitational acceleration. Since the student’s upward flight is exactly matched by his or her downward flight. Because the flight is matched, the student spends $t\/2$ time rising and $t\/2$ time descending. Since the student has no vertical speed at the top, we know that his or her speed is
\n$$\u00a0 s = g\\frac{t}{2},$$<\/p>\n

where $g$ is the gravitational acceleration 9.8 m\/s2<\/sup>.<\/p>\n

Now, we have a system of equations<\/p>\n

$$\u00a0 d = t\\cdot s $$
\n$$\u00a0 s = g\\frac{t}{2}.$$
\nThe system looks like it has a many variables, but really there are only two, $s$ and $t$. We know $g$ and $d$. To solve the system we substitute for $s$ in the first equation with the second to get<\/p>\n

$$\u00a0 d = tg\\frac{t}{2} = g\\frac{t^2}{2}$$
\nSolve for $t$
\n$$\u00a0\u00a0 t = \\sqrt{\\frac{2d}{g}}
\n= \\sqrt{\\frac{2\\cdot 5.73\\,\\text{m}}{9.8\\,\\text{m\/s}^2}}
\n\\approx 34.2\\,\\text{s}.
\n$$
\nNot a bad commute, a little over half a minute.<\/p>\n

With $t$ in hand, we can find the magnitude of the initial velocity. Remember that the initial velocity is $s$ in the horizontal direction and $s$ in the vertical direction, so the speed when leaving the launcher is
\n$$
\n\\left| \\mathbf{v}_0\\right| = \\sqrt{s^2 + s^2} = \\sqrt{2s^2} = s\\sqrt{2}.
\n$$
\nThe initial speed the student must attain is given by the very first equation, $d = s\\cdot t$. Solving for $s$ with the value of $t$ we found, we get<\/p>\n

$$\u00a0 s = \\frac{5.73\\,\\text{km}}{34.2\\,\\text{s}} = 168\\,\\text{m\/s}. $$<\/p>\n

Finding the Acceleration<\/h2>\n

The ramp lives on a footprint that is about 80 ft, or 24.4 m. It is also 24.4 m tall, so special zoning is surely required! The rail of the launch-o-rocket is the hypotenuse of a triangle, and that triangle has sides 24.4 m, and a total length of $\\sqrt{2}\\cdot 24.4\\,\\text{m} = 34.5\\,\\text{m}$.<\/p>\n

The formula for position after a period of acceleration is
\n$$\u00a0\u00a0 p = \\frac{1}{2}a\\tau^2.$$
\nFor our system, we also know that the acceleration is the change in speed divided by the change in time. Our speed goes from zero to 168\u00a0m\/s in $\\tau$. Again, we have a system of equations,<\/p>\n

$$\u00a0 34.5\\, \\text{m} = \\frac{1}{2}a\\tau^2 $$
\n$$\u00a0 a = \\frac{168\\,\\text{m\/s}}{\\tau}.$$<\/p>\n

Solve for $a$ by first solving the second equation for $\\tau$, and then substituting that result into the first equation to get<\/p>\n

$$\u00a0\u00a0 34.5\\, \\text{m} = \\frac{1}{2}a\\left(\\frac{168\\,\\text{m\/s}}{a}\\right)^2 $$<\/p>\n

$$\u00a0\u00a0 a = \\frac{\\left( 168\\, \\text{m\/s}\\right)^2}{2 \\cdot 34.5\\,\\text{m}}
\n= 407\\, \\text{m\/s}^2 = 41.5\\, g. $$<\/p>\n

The typical onset of death occurs when acceleration exceeds about $10g$, so unfortunately, the launch-o-rocket is a single try system.<\/em><\/p>\n","protected":false},"excerpt":{"rendered":"

School, from our house as the crow flies, is 5.73 km. If we neglect air resistance and deal strictly with ballistic flight then we can materialize a wonderful fantasy. Starting in the backyard, extending over the top of the house, is a launch-o-rocket, a rail-like launcher that accelerates the school-bound student until he or she […]<\/p>\n","protected":false},"author":1,"featured_media":1094,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-1095","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/www.inkofpark.com\/index.php?rest_route=\/wp\/v2\/posts\/1095","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.inkofpark.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.inkofpark.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.inkofpark.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.inkofpark.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1095"}],"version-history":[{"count":11,"href":"https:\/\/www.inkofpark.com\/index.php?rest_route=\/wp\/v2\/posts\/1095\/revisions"}],"predecessor-version":[{"id":1382,"href":"https:\/\/www.inkofpark.com\/index.php?rest_route=\/wp\/v2\/posts\/1095\/revisions\/1382"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.inkofpark.com\/index.php?rest_route=\/wp\/v2\/media\/1094"}],"wp:attachment":[{"href":"https:\/\/www.inkofpark.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1095"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.inkofpark.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1095"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.inkofpark.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1095"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}