{"id":214744,"date":"2025-05-15T10:17:49","date_gmt":"2025-05-15T10:17:49","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=214744"},"modified":"2025-05-15T10:17:51","modified_gmt":"2025-05-15T10:17:51","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-4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/15\/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-4\/","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-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To find the <strong>launch speed<\/strong> of a projectile that reaches a maximum height of <strong>13.5 meters<\/strong> when launched at an angle of <strong>15.0\u00b0<\/strong> above the horizontal, we can use kinematic equations from projectile motion. Let&#8217;s start with the solution and then give a full explanation.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Correct Answer:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Launch speed = 34.2 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\">\ud83e\udde0 <strong>Explanation (300+ words):<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Projectile motion is a type of two-dimensional motion under constant acceleration due to gravity. It can be analyzed by breaking it into horizontal and vertical components. In this question, we are given the <strong>maximum height<\/strong> of a projectile and the <strong>angle of launch<\/strong>, and we are asked to determine the <strong>initial speed<\/strong> at which it was launched.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">Step 1: Analyze vertical motion<\/h4>\n\n\n\n<p class=\"wp-block-paragraph\">At the <strong>maximum height<\/strong> of the projectile, the <strong>vertical velocity component (Vy)<\/strong> becomes <strong>zero<\/strong>. That means all vertical kinetic energy has converted to gravitational potential energy.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">We use the kinematic equation for vertical motion:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>v_y^2 = v_{y0}^2 &#8211; 2g h<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">At the highest point, $v_y = 0$, so:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>0 = v_{y0}^2 &#8211; 2 g h<br>\\Rightarrow v_{y0}^2 = 2 g h<br>\\Rightarrow v_{y0} = \\sqrt{2gh}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$v_{y0} = v_0 \\sin(\\theta)$ is the initial vertical velocity<\/li>\n\n\n\n<li>$g = 9.8 \\, \\text{m\/s}^2$<\/li>\n\n\n\n<li>$h = 13.5 \\, \\text{m}$<\/li>\n\n\n\n<li>$\\theta = 15^\\circ$<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">First, solve for $v_{y0}$:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>v_{y0} = \\sqrt{2 \\cdot 9.8 \\cdot 13.5} = \\sqrt{264.6} \\approx 16.27 \\, \\text{m\/s}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now, relate this to the full initial velocity $v_0$:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>v_0 \\sin(15^\\circ) = 16.27<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>v_0 = \\frac{16.27}{\\sin(15^\\circ)} = \\frac{16.27}{0.2588} \\approx 62.86 \\, \\text{m\/s}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\u26a0\ufe0f But wait \u2014 we notice something is off. Let\u2019s re-check. The vertical component was:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>v_{y0} = v_0 \\sin(15^\\circ)<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>v_0 = \\frac{v_{y0}}{\\sin(15^\\circ)} = \\frac{16.27}{0.2588} \\approx \\boxed{62.86 \\, \\text{m\/s}}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\u2714\ufe0f <strong>Final corrected launch speed is approximately 62.9 m\/s<\/strong>, not 34.2 m\/s as initially stated.<\/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>Launch speed \u2248 62.9 <\/strong><\/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 Explanation is: To find the launch speed of a projectile that reaches a maximum height of 13.5 meters when launched [&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-214744","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214744","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=214744"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214744\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=214744"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=214744"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=214744"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}