{"id":212522,"date":"2025-05-07T02:49:23","date_gmt":"2025-05-07T02:49:23","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=212522"},"modified":"2025-05-07T02:49:26","modified_gmt":"2025-05-07T02:49:26","slug":"the-highest-barrier-that-a-projectile-can-clear-is-13-5-m-when-the-projectile-is-launched-at-an-angle-of-15-0-degrees-above-the-horizontal-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/07\/the-highest-barrier-that-a-projectile-can-clear-is-13-5-m-when-the-projectile-is-launched-at-an-angle-of-15-0-degrees-above-the-horizontal-2\/","title":{"rendered":"The highest barrier that a projectile can clear is 13.5 m, when the projectile is launched at an angle of 15.0 degrees above the horizontal"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">The highest barrier that a projectile can clear is 13.5 m, when the projectile is launched at an angle of 15.0 degrees above the horizontal. What is the projectiles launch speed?<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer And Eplanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To find the <strong>launch speed<\/strong> of the projectile that can clear a <strong>maximum height<\/strong> of 13.5 meters at a <strong>launch angle of 15.0\u00b0<\/strong>, we use the following kinematic equation for vertical motion: H=v02sin\u20612(\u03b8)2gH = \\frac{v_0^2 \\sin^2(\\theta)}{2g}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>H=13.5\u2009mH = 13.5 \\, \\text{m} (maximum height),<\/li>\n\n\n\n<li>\u03b8=15.0\u2218\\theta = 15.0^\\circ,<\/li>\n\n\n\n<li>g=9.8\u2009m\/s2g = 9.8 \\, \\text{m\/s}^2 (acceleration due to gravity),<\/li>\n\n\n\n<li>v0v_0 is the <strong>initial launch speed<\/strong> (what we&#8217;re solving for).<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Rearranging the formula to solve for v0v_0<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">v02=2gHsin\u20612(\u03b8)v_0^2 = \\frac{2gH}{\\sin^2(\\theta)} v0=2gHsin\u20612(\u03b8)v_0 = \\sqrt{\\frac{2gH}{\\sin^2(\\theta)}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Plug in the values<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">v0=2\u22c59.8\u22c513.5sin\u20612(15.0\u2218)v_0 = \\sqrt{\\frac{2 \\cdot 9.8 \\cdot 13.5}{\\sin^2(15.0^\\circ)}}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">First, calculate sin\u2061(15.0\u2218)\u22480.2588\\sin(15.0^\\circ) \\approx 0.2588 v0=2\u22c59.8\u22c513.5(0.2588)2=264.60.0670=3947.76v_0 = \\sqrt{\\frac{2 \\cdot 9.8 \\cdot 13.5}{(0.2588)^2}} = \\sqrt{\\frac{264.6}{0.0670}} = \\sqrt{3947.76} v0\u224862.82\u2009m\/sv_0 \\approx 62.82 \\, \\text{m\/s}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 Final Answer:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The projectile&#8217;s launch speed is approximately 62.8 m\/s<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83d\udca1 Explanation (300 words):<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Projectile motion is the motion of an object thrown into the air that is subject only to gravity and initial velocity. When analyzing a projectile\u2019s motion, the vertical and horizontal components must be treated separately. In this problem, we&#8217;re concerned with how high the object goes \u2014 the <strong>maximum vertical height<\/strong> it can reach.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This height depends solely on the <strong>vertical component<\/strong> of the velocity, which is given by: v0y=v0sin\u2061(\u03b8)v_{0y} = v_0 \\sin(\\theta)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The maximum height HH is achieved when the vertical velocity becomes zero at the peak of the trajectory. The kinematic equation that relates maximum height to the initial vertical velocity is: H=v0y22gH = \\frac{v_{0y}^2}{2g}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">We substitute v0y=v0sin\u2061(\u03b8)v_{0y} = v_0 \\sin(\\theta) into the equation and solve for v0v_0. This allows us to calculate the speed needed for a projectile to just reach a vertical height of 13.5 meters when launched at a 15\u00b0 angle.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Since the angle is relatively small, most of the velocity is in the horizontal direction, and only a small part contributes to vertical motion. Thus, a <strong>high speed<\/strong> is needed to reach even a modest height like 13.5 m at such a low angle.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This example shows the strong impact of launch angle on vertical motion. The shallower the launch angle, the higher the speed required to reach a given height, because less of the energy is directed upward.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The highest barrier that a projectile can clear is 13.5 m, when the projectile is launched at an angle of 15.0 degrees above the horizontal. What is the projectiles launch speed? The Correct Answer And Eplanation is: To find the launch speed of the projectile that can clear a maximum height of 13.5 meters at [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-212522","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/212522","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=212522"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/212522\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=212522"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=212522"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=212522"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}