{"id":1867,"date":"2024-06-14T21:38:09","date_gmt":"2024-06-14T21:38:09","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=1867"},"modified":"2025-08-14T01:14:23","modified_gmt":"2025-08-14T01:14:23","slug":"analysis-of-quadratic-functions-apply-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/analysis-of-quadratic-functions-apply-it-1\/","title":{"raw":"Analysis of Quadratic Functions: Apply It 1","rendered":"Analysis of Quadratic Functions: Apply It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n \t<li>Use quadratic equations to figure out solutions to real-life situations<\/li>\r\n<\/ul>\r\n<\/section>\r\n\r\n[caption id=\"attachment_3360\" align=\"alignleft\" width=\"440\"]<img class=\"wp-image-3360\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/10181832\/image19-300x225.jpg\" alt=\"\" width=\"440\" height=\"330\" \/> Trebuchet sitting idle[\/caption]\r\n\r\nNow that you have seen how to work with quadratic functions in this module, let\u2019s apply our skills to a problem involving projectile motion.\r\n\r\nA <strong>trebuchet<\/strong> is a medieval weapon used to hurl large stones at enemy forts or castles in order to smash through the thick walls. \u00a0Basically the trebuchet works by putting the stone in a sling at the end of a long pole which is allowed to rotate around a fixed axis. \u00a0The shorter end of the pole is attached to a very heavy weight to counterbalance the projectile. \u00a0When the weight is released at the short end, the pole rotates, flinging the stone at high speed upward and toward the enemy.\r\n\r\n<section class=\"textbox example\">Suppose a particular trebuchet can launch a [latex]10[\/latex] kilogram stone with an initial upward speed of [latex]24.5[\/latex] meters per second. Furthermore, suppose that the stone is [latex]8[\/latex] meters above the ground when it leaves the sling \u2013 that is, its initial height is [latex]8[\/latex] m.We can use what we know about quadratic functions to answer some interesting questions about the stone that is launched from the trebuchet:\r\n<ol>\r\n \t<li style=\"font-weight: 400;\">What is the maximum height achieved by the stone during its flight to the enemy castle?<\/li>\r\n \t<li style=\"font-weight: 400;\">When does that maximum occur?<\/li>\r\n \t<li style=\"font-weight: 400;\">How long would it take before the stone comes crashing down to Earth (in case it missed the castle altogether)?<\/li>\r\n<\/ol>\r\nIn order to answer these questions, we need to know about a particular quadratic function from physics called the <strong>ballistics equation<\/strong>:\r\n<p style=\"text-align: center;\">[latex]h(t)=-{\\large\\frac{g}{2}}t^2+v_0t+h_0[\/latex]<\/p>\r\nHere, [latex]h(t)[\/latex] stands for the height at time [latex]t[\/latex], [latex]v_0[\/latex] is the initial upward velocity (speed), [latex]h_0[\/latex] is the initial height of the projectile, and [latex]g[\/latex] is a constant called the acceleration due to gravity. \u00a0Near the surface of the Earth, [latex]g\\approx9.8[\/latex] meters per second squared, so [latex]-{\\Large\\frac{g}{2}}\\approx-4.9[\/latex].\r\n\r\nNow putting the given values into their proper places, we have:\r\n<p style=\"text-align: center;\">[latex]h(t)=-4.9t^2+24.5t+8[\/latex]<\/p>\r\nThis is a quadratic function with [latex]a=-4.9[\/latex], [latex]b=24.5[\/latex], [latex]c=8[\/latex]. \u00a0The graph is a parabola, and it will have a <span style=\"text-decoration: underline;\">maximum<\/span> because [latex]a = -4.9 &lt; 0[\/latex]. \u00a0The maximum function value in this case represents the maximum height of the stone.\r\n<p style=\"text-align: left;\">Now we can answer the first two questions:<\/p>\r\n\r\n<ul>\r\n \t<li style=\"text-align: left;\">What is the maximum height achieved by the stone during its flight to the enemy castle?<\/li>\r\n \t<li>When does that maximum occur?<\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 1x;\">To find the maximum height and time at which it occurs, we use the vertex formula which will give [latex](t_\\text{max},h(t_\\text{max}))[\/latex]. the time at maximum point and height at the time of the maximum point.<\/p>\r\n<p style=\"text-align: center;\">[latex]t_\\text{max}=-{\\large\\frac{b}{2a}}=-{\\large\\frac{24.5}{2(-4.9)}}=2.5[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]h(t_\\text{max})=h\\left(-{\\large\\frac{b}{2a}}\\right)=h(2.5)=-4.9(2.5)^2+24.5(2.5)+8\\approx38.6[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 40px;\"><strong>Therefore, the maximum height will be 38.6 meters, which occurs 2.5 seconds after launch. \u00a0That\u2019s a high-flying projectile!<\/strong><\/p>\r\n<p style=\"text-align: left;\">Finally, to determine when the stone hits the ground, we just have to find the [latex]x[\/latex]-intercept of the function. \u00a0In other words, we have to solve [latex]h(t)=0[\/latex]. \u00a0That\u2019s a job for the quadratic formula.<\/p>\r\n<p style=\"text-align: center;\">[latex]t_\\text{final}={\\large\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}}={\\large\\frac{-24.5\\pm\\sqrt{24.5^2-4(4.9)(8)}}{2a}}\\approx-0.3,5.3[\/latex]<\/p>\r\n<p style=\"text-align: left;\">As expected, the quadratic formula gives us two answers, but only the positive one makes sense in this context.<\/p>\r\n<p style=\"text-align: left;\"><strong>The stone lands on the ground about 5.3 seconds after it was launched.<\/strong><\/p>\r\n\r\n<\/section>Quadratic functions can be used to model the behavior of objects in free fall, amongst other things. We can use algebra to analyze this behavior for interesting features.\r\n\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]24490[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Use quadratic equations to figure out solutions to real-life situations<\/li>\n<\/ul>\n<\/section>\n<figure id=\"attachment_3360\" aria-describedby=\"caption-attachment-3360\" style=\"width: 440px\" class=\"wp-caption alignleft\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-3360\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/10181832\/image19-300x225.jpg\" alt=\"\" width=\"440\" height=\"330\" \/><figcaption id=\"caption-attachment-3360\" class=\"wp-caption-text\">Trebuchet sitting idle<\/figcaption><\/figure>\n<p>Now that you have seen how to work with quadratic functions in this module, let\u2019s apply our skills to a problem involving projectile motion.<\/p>\n<p>A <strong>trebuchet<\/strong> is a medieval weapon used to hurl large stones at enemy forts or castles in order to smash through the thick walls. \u00a0Basically the trebuchet works by putting the stone in a sling at the end of a long pole which is allowed to rotate around a fixed axis. \u00a0The shorter end of the pole is attached to a very heavy weight to counterbalance the projectile. \u00a0When the weight is released at the short end, the pole rotates, flinging the stone at high speed upward and toward the enemy.<\/p>\n<section class=\"textbox example\">Suppose a particular trebuchet can launch a [latex]10[\/latex] kilogram stone with an initial upward speed of [latex]24.5[\/latex] meters per second. Furthermore, suppose that the stone is [latex]8[\/latex] meters above the ground when it leaves the sling \u2013 that is, its initial height is [latex]8[\/latex] m.We can use what we know about quadratic functions to answer some interesting questions about the stone that is launched from the trebuchet:<\/p>\n<ol>\n<li style=\"font-weight: 400;\">What is the maximum height achieved by the stone during its flight to the enemy castle?<\/li>\n<li style=\"font-weight: 400;\">When does that maximum occur?<\/li>\n<li style=\"font-weight: 400;\">How long would it take before the stone comes crashing down to Earth (in case it missed the castle altogether)?<\/li>\n<\/ol>\n<p>In order to answer these questions, we need to know about a particular quadratic function from physics called the <strong>ballistics equation<\/strong>:<\/p>\n<p style=\"text-align: center;\">[latex]h(t)=-{\\large\\frac{g}{2}}t^2+v_0t+h_0[\/latex]<\/p>\n<p>Here, [latex]h(t)[\/latex] stands for the height at time [latex]t[\/latex], [latex]v_0[\/latex] is the initial upward velocity (speed), [latex]h_0[\/latex] is the initial height of the projectile, and [latex]g[\/latex] is a constant called the acceleration due to gravity. \u00a0Near the surface of the Earth, [latex]g\\approx9.8[\/latex] meters per second squared, so [latex]-{\\Large\\frac{g}{2}}\\approx-4.9[\/latex].<\/p>\n<p>Now putting the given values into their proper places, we have:<\/p>\n<p style=\"text-align: center;\">[latex]h(t)=-4.9t^2+24.5t+8[\/latex]<\/p>\n<p>This is a quadratic function with [latex]a=-4.9[\/latex], [latex]b=24.5[\/latex], [latex]c=8[\/latex]. \u00a0The graph is a parabola, and it will have a <span style=\"text-decoration: underline;\">maximum<\/span> because [latex]a = -4.9 < 0[\/latex]. \u00a0The maximum function value in this case represents the maximum height of the stone.\n\n\n<p style=\"text-align: left;\">Now we can answer the first two questions:<\/p>\n<ul>\n<li style=\"text-align: left;\">What is the maximum height achieved by the stone during its flight to the enemy castle?<\/li>\n<li>When does that maximum occur?<\/li>\n<\/ul>\n<p style=\"padding-left: 1x;\">To find the maximum height and time at which it occurs, we use the vertex formula which will give [latex](t_\\text{max},h(t_\\text{max}))[\/latex]. the time at maximum point and height at the time of the maximum point.<\/p>\n<p style=\"text-align: center;\">[latex]t_\\text{max}=-{\\large\\frac{b}{2a}}=-{\\large\\frac{24.5}{2(-4.9)}}=2.5[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]h(t_\\text{max})=h\\left(-{\\large\\frac{b}{2a}}\\right)=h(2.5)=-4.9(2.5)^2+24.5(2.5)+8\\approx38.6[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 40px;\"><strong>Therefore, the maximum height will be 38.6 meters, which occurs 2.5 seconds after launch. \u00a0That\u2019s a high-flying projectile!<\/strong><\/p>\n<p style=\"text-align: left;\">Finally, to determine when the stone hits the ground, we just have to find the [latex]x[\/latex]-intercept of the function. \u00a0In other words, we have to solve [latex]h(t)=0[\/latex]. \u00a0That\u2019s a job for the quadratic formula.<\/p>\n<p style=\"text-align: center;\">[latex]t_\\text{final}={\\large\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}}={\\large\\frac{-24.5\\pm\\sqrt{24.5^2-4(4.9)(8)}}{2a}}\\approx-0.3,5.3[\/latex]<\/p>\n<p style=\"text-align: left;\">As expected, the quadratic formula gives us two answers, but only the positive one makes sense in this context.<\/p>\n<p style=\"text-align: left;\"><strong>The stone lands on the ground about 5.3 seconds after it was launched.<\/strong><\/p>\n<\/section>\n<p>Quadratic functions can be used to model the behavior of objects in free fall, amongst other things. We can use algebra to analyze this behavior for interesting features.<\/p>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm24490\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24490&theme=lumen&iframe_resize_id=ohm24490&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":12,"menu_order":21,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":185,"module-header":"apply_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1867"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":7,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1867\/revisions"}],"predecessor-version":[{"id":7735,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1867\/revisions\/7735"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/185"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1867\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=1867"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1867"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=1867"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=1867"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}