{"id":790,"date":"2020-12-09T14:49:27","date_gmt":"2020-12-09T14:49:27","guid":{"rendered":"https:\/\/pickedshares.com\/?p=790"},"modified":"2021-05-08T12:15:11","modified_gmt":"2021-05-08T12:15:11","slug":"engineering-mechanics-3-exercise-3-straight-throw-determination-of-the-initial-speed","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-3-exercise-3-straight-throw-determination-of-the-initial-speed\/","title":{"rendered":"Straight throw, determination of the initial speed"},"content":{"rendered":"\n<h3 class=\"wp-block-heading\">Task<\/h3>\n\n\n\n<p>A lorry fell into a shaft and left traces on the wall at depth H under the crash site. At what speed did the lorry go into the shaft? (The air resistance can be neglected.)<\/p>\n\n\n\n<p>Given: H = 20 m, b = 4,5 m, g = 9,81 m\/s\u00b2<\/p>\n\n\n\n<figure class=\"wp-block-image size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201122-112608_Sketch-834x1024.jpg\" alt=\"Determination of the initial speed, straight throw\" class=\"wp-image-475\" width=\"480\" height=\"589\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201122-112608_Sketch-834x1024.jpg 834w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201122-112608_Sketch-244x300.jpg 244w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201122-112608_Sketch-768x943.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201122-112608_Sketch-400x491.jpg 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201122-112608_Sketch-800x982.jpg 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201122-112608_Sketch.jpg 1142w\" sizes=\"auto, (max-width: 480px) 100vw, 480px\" \/><figcaption>Determination of the initial speed, straight 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 acceleration is<\/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<p>\\[\\tag{1} \\ddot{x} = 0 \\]<\/p>\n<p>\\[\\tag{2} \\ddot{z} = -g \\]<\/p>\n\n\n\n<p><\/p>\n\n\n\n<p>The functions in the x and z directions for speed and position are obtained through double integration.<\/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\n<p>\\[ \\tag{3} \\dot{x} = c_1  \\]<\/p>\n<p>\\[ \\tag{4} x(t) = c_1t+c_3  \\]<\/p>\n\n<p>\\[ \\tag{5} \\dot{z}(t) = -gt + c_2 \\]<\/p>\n<p>\\[ \\tag{6} z(t) = -\\frac{1}{2}gt^2 + c_2t + c_4  \\]<\/p>\n\n\n\n<p>The speed at time t = 0<\/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\n<p>\\[ \\tag{7} \\dot{\\vec{x}}(t = 0) = v_0  \\]<\/p>\n<p>\\[ \\tag{8} \\dot{\\vec{z}}(t = 0) =  0 \\]<\/p>\n\n\n\n<p>which means that c<sub>1<\/sub> and c<sub>2<\/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\n<p>\\[ \\tag{9} v_0  = c_1  \\]<\/p>\n<p>\\[ \\tag{10} c_1 = v_0   \\]<\/p>\n<p>\\[ \\tag{11}  0 =  \\bcancel{-gt} + c_2   \\]<\/p>\n<p>\\[ \\tag{12} c_2 = 0   \\]<\/p>\n\n\n\n<p>From the initial conditions for the position at time t = 0, c<sub>3<\/sub> and c<sub>4<\/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\n<p>\\[ \\tag{13} \\vec{x}(t = 0) = 0  \\]<\/p>\n<p>\\[ \\tag{14} 0 =  \\bcancel{v_0t}+c_3  \\]<\/p>\n<p>\\[ \\tag{15} c_3 = 0  \\]<\/p>\n\n<p>\\[ \\tag{16} \\vec{z}(t = 0) = 0  \\]<\/p>\n<p>\\[ \\tag{17}  0 = \\bcancel{-\\frac{1}{2}gt^2} + c_4  \\]<\/p>\n<p>\\[ \\tag{18} c_4 = 0  \\]<\/p>\n\n\n\n<p>The as yet unknown point in time of the impact is now referred to as T. The position of the cart at time T is:<\/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} \\vec{x}(t = T) = b  \\]<\/p>\n<p>\\[ \\tag{20} \\vec{z}(t = T) = -H \\]<\/p>\n\n\n\n<p>From which both the time T and the initial speed v<sub>0<\/sub> can be calculated in the following.<\/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{21} b  =  v_0T  \\]<\/p>\n<p>\\[ \\tag{22}  -H =  -\\frac{1}{2}gT^2  \\]<\/p>\n<p>\\[ \\tag{23} T = \\sqrt{\\frac{2H}{g}}  \\]<\/p>\n<p>\\[ \\tag{24} v_0 = \\frac{b}{\\sqrt{\\frac{2H}{g}}}  \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{25} v_0 = \\frac{ 4.5 m}{ \\sqrt{ \\frac{ 40 m}{ 9.81 \\frac{ m }{ s^2 }}}} = 2.23 \\frac{ m }{ s }  \\]<\/p>\n<\/div>\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Straight throw, determination of the initial speed\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-3-exercise-3-straight-throw-determination-of-the-initial-speed\/\" aria-label=\"Read more about Straight throw, determination of the initial speed\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":478,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[34,60],"tags":[25],"class_list":["post-790","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>Straight throw, determination of the initial speed &#8226; pickedshares<\/title>\n<meta name=\"description\" content=\"A lorry has been fallen into a shaft. 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