{"id":793,"date":"2020-12-09T16:43:42","date_gmt":"2020-12-09T16:43:42","guid":{"rendered":"https:\/\/pickedshares.com\/?p=793"},"modified":"2021-05-08T12:16:57","modified_gmt":"2021-05-08T12:16:57","slug":"engineering-mechanics-3-exercise-4-shot-at-a-raised-target","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-3-exercise-4-shot-at-a-raised-target\/","title":{"rendered":"Shot at a raised target"},"content":{"rendered":"\n<h3 class=\"wp-block-heading\">Task<\/h3>\n\n\n\n<p>A gun is intended to hit an elevated target at a given angle. How big does the muzzle velocity v<sub>0<\/sub> have to be? What is the minimum size of the angle for the present case? (The air resistance and the Coriolis force can be neglected.)<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"532\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-214640_Sketch-1024x532.jpg\" alt=\"Ballistic trajectory, angled throw\" class=\"wp-image-480\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-214640_Sketch-1024x532.jpg 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-214640_Sketch-300x156.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-214640_Sketch-768x399.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-214640_Sketch-400x208.jpg 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-214640_Sketch-800x416.jpg 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-214640_Sketch.jpg 1383w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Ballistic trajectory, angled throw<\/figcaption><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\">Solution<\/h3>\n\n\n\n<p>A solution video will be published soon <a href=\"http:\/\/www.youtube.com\/channel\/UCSh8qo2-6dhtbZqSLStt3Rg\" target=\"_blank\" rel=\"noreferrer noopener\">on this channel<\/a>.<\/p>\n\n\n\n<p>The accelerations in direction x and direction y are<\/p>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\n<p>\\[ \\tag{1} \\ddot{x} = 0 \\]<\/p>\n<p>\\[ \\tag{2} \\ddot{z} = -g \\]<\/p>\n\n\n\n<p>The integration of both accelerations leads to the velocity<\/p>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\n<p>\\[ \\tag{3} \\dot{x} = c_1 \\]<\/p>\n<p>\\[ \\tag{4} \\dot{z}(t) = -gt + c_2 \\]<\/p>\n\n\n\n<p>and the location.<\/p>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n<p>\\[ \\tag{5} x(t) = c_1t + c_3 \\]<\/p>\n<p>\\[ \\tag{6} z(t) = -\\frac{1}{2}gt^2 + c_2t + c_4\\]<\/p>\n\n\n\n<p>The position of the projectile at time t = 0 in the x-direction is equal to zero, from which c<sub>3<\/sub> can be calculated.<\/p>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n<p>\\[ \\tag{7} x(t=0) = 0 \\]<\/p>\n<p>\\[ \\tag{8} 0 = \\bcancel{c_1t} + c_3 \\]<\/p>\n<p>\\[ \\tag{9} c_3 = 0 \\]<\/p>\n\n\n\n<p>The z-position at time t = 0 is also zero, which results in c<sub>4<\/sub>.<\/p>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\n<p>\\[ \\tag{10} z(t=0) = 0 \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{11} 0 = \\bcancel{-\\frac{1}{2}gt^2} + \\bcancel{c_2t} + c_4 \\]<\/p>\n<\/div>\n<p>\\[ \\tag{12} c_4 = 0 \\]<\/p>\n\n\n\n<p>The speed in the x direction at time t = 0 is the horizontal component of v<sub>0<\/sub>, which delivers c<sub>1<\/sub>.<\/p>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n<p>\\[ \\tag{13} \\dot{x}(t=0) = v_0 \\cdot cos \\alpha \\]<\/p>\n<p>\\[ \\tag{14} c_1 = v_0 \\cdot cos \\alpha \\]<\/p>\n\n<p>\\[ \\tag{15} \\dot{x} = v_0 \\cdot cos \\alpha \\]<\/p>\n\n\n\n<p>The speed in the z-direction is the vertical component of v<sub>0<\/sub>, so that c<sub>2<\/sub> can be determined.<\/p>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n<p>\\[ \\tag{16} \\dot{z}(t=0) = v_0 \\cdot sin \\alpha \\]<\/p>\n<p>\\[ \\tag{17} v_0 \\cdot sin \\alpha = \\bcancel{-gt} + c_2 \\]<\/p>\n<p>\\[ \\tag{18} c_2 = v_0 \\cdot sin \\alpha \\]<\/p>\n\n\n\n<p>With the integration constants determined so far, the functions for the position of the projectile now look like this:<\/p>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\n<p>\\[ \\tag{19} x(t) = v_0 \\cdot cos \\alpha \\cdot t \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{20} z(t) = -\\frac{1}{2}gt^2 + v_0 \\cdot sin \\alpha \\cdot t\\]<\/p>\n<\/div>\n\n\n\n<p>The time of impact is referred to as T in the following. The x position at time t = T is 2a.<\/p>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\n<p>\\[ \\tag{21} x(t=T) = 2a \\]<\/p>\n<p>\\[ \\tag{22} 2a = v_0 \\cdot cos \\alpha \\cdot T \\]<\/p>\n<p>\\[ \\tag{23} T = \\frac{2a}{v_0 \\cdot cos \\alpha} \\]<\/p>\n\n\n\n<p>The z-position at time T is a.<\/p>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\n<p>\\[ \\tag{24} z(t=T) = a\\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{25} a = -\\frac{1}{2}gT^2 + v_0 \\cdot sin \\alpha \\cdot T\\]<\/p>\n<\/div>\n\n\n\n<p>The calculated T (equation 23) is now inserted and the equation can be solved for v<sub>0<\/sub>.<\/p>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{26} a = -\\frac{1}{2}g \\cdot \\left( \\frac{2a}{v_0 \\cdot cos \\alpha} \\right)^2 + v_0 \\cdot sin \\alpha \\cdot \\frac{2a}{v_0 \\cdot cos \\alpha}\\]<\/p>\n<p>\\[ \\tag{27} a = -\\frac{1}{2}g \\cdot \\frac{4a^2}{v_0^2 \\cdot cos^2 \\alpha} + sin \\alpha \\cdot \\frac{2a}{cos \\alpha}\\]<\/p>\n<p>\\[ \\tag{28} a = -\\frac{1}{2}g \\cdot \\frac{4a^2}{v_0^2 \\cdot (cos(2\\alpha)+1) } + 2a \\cdot tan \\alpha\\]<\/p>\n<p>\\[ \\tag{29} 4a \\cdot tan \\alpha -2a = g \\cdot \\frac{4a^2}{v_0^2 \\cdot (cos(2\\alpha)+1) } \\]<\/p>\n<p>\\[ \\tag{30} \\frac{(4a \\cdot tan \\alpha -2a)\\cdot (cos(2\\alpha)+1)}{4a^2g} = \\frac{1}{v_0^2  } \\]<\/p>\n<p>\\[ \\tag{31} v_0 = \\sqrt{\\frac{4a^2g}{(4a \\cdot tan \\alpha -2a)\\cdot (cos(2\\alpha)+1)}}   \\]<\/p>\n<\/div>\n\n\n\n<h4 class=\"wp-block-heading\">Determination of the minimum angle<\/h4>\n\n\n\n<p>Consider the two terms in brackets below the brackets in equation 31. If one of the two becomes zero, the equation no longer provides a solution. So it applies<\/p>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n<p>\\[ \\tag{32} 4a \\cdot tan \\alpha -2a &gt; 0   \\]<\/p>\n<p>\\[ \\tag{33} \\alpha &gt; arctan \\frac{1}{2}   \\]<\/p>\n<p>\\[ \\tag{34} \\alpha &gt; 26.57\u00b0   \\]<\/p>\n\n\n\n<p>as<\/p>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n<p>\\[ \\tag{35} cos(2\\alpha)+1 &gt; 0   \\]<\/p>\n<p>\\[ \\tag{36} \\alpha &gt; \\frac{arccos(-1)}{2}   \\]<\/p>\n<p>\\[ \\tag{37} \\alpha &gt; 180\u00b0   \\]<\/p>\n\n\n\n<p>Obviously, the minimum angle is found with equation 34, while equation 37 represents the upper limit value.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Validation of the function<\/h4>\n\n\n\n<p>Below are two diagram plots with the x and y coordinates of the projectile for two different angles of fire (45 \u00b0 and 60 \u00b0) with the same parameters for distance and height. The destination is marked with a cross.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1017\" height=\"719\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-215030_Sketch.jpg\" alt=\"Shot under 45\u00b0\" class=\"wp-image-481\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-215030_Sketch.jpg 1017w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-215030_Sketch-300x212.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-215030_Sketch-768x543.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-215030_Sketch-400x283.jpg 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-215030_Sketch-800x566.jpg 800w\" sizes=\"auto, (max-width: 1017px) 100vw, 1017px\" \/><figcaption>Shot under 45\u00b0<\/figcaption><\/figure>\n\n\n\n<p>At a firing angle of 45 \u00b0, the projectile hits the target at the apex of its trajectory.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"984\" height=\"716\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-215209_Sketch.jpg\" alt=\"Shot under 60 \u00b0\" class=\"wp-image-482\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-215209_Sketch.jpg 984w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-215209_Sketch-300x218.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-215209_Sketch-768x559.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-215209_Sketch-400x291.jpg 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201123-215209_Sketch-800x582.jpg 800w\" sizes=\"auto, (max-width: 984px) 100vw, 984px\" \/><figcaption>Shot under 60 \u00b0<\/figcaption><\/figure>\n\n\n\n<p>At a firing angle of 60 \u00b0, the projectile moves on a significantly elevated trajectory and hits the target in its downward movement.<\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Shot at a raised target\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-3-exercise-4-shot-at-a-raised-target\/\" aria-label=\"Read more about Shot at a raised target\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":479,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[34,60],"tags":[25],"class_list":["post-793","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-engineering-mechanics-iii","category-exercises","tag-kinematics","infinite-scroll-item","masonry-post","generate-columns","tablet-grid-50","mobile-grid-100","grid-parent","grid-33"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Shot at a raised target &#8226; pickedshares<\/title>\n<meta name=\"description\" content=\"A cannon shall be fired at a raised target under a given angle. 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