{"id":213247,"date":"2025-05-10T05:36:18","date_gmt":"2025-05-10T05:36:18","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=213247"},"modified":"2025-05-10T05:36:21","modified_gmt":"2025-05-10T05:36:21","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-3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/10\/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-3\/","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<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To determine the <strong>launch speed<\/strong> of a projectile that clears a barrier of <strong>13.5 m<\/strong> at a launch angle of <strong>15.0\u00b0<\/strong>, we need to analyze the <strong>vertical motion<\/strong> using <strong>kinematics<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Use vertical motion formula<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The formula for the <strong>maximum height<\/strong> $h$ of a projectile is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>h = \\frac{v_0^2 \\sin^2\\theta}{2g}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$h = 13.5 \\, \\text{m}$ (maximum height),<\/li>\n\n\n\n<li>$v_0$ = launch speed (unknown),<\/li>\n\n\n\n<li>$\\theta = 15.0^\\circ$,<\/li>\n\n\n\n<li>$g = 9.8 \\, \\text{m\/s}^2$ (acceleration due to gravity),<\/li>\n\n\n\n<li>$\\sin^2\\theta = (\\sin 15^\\circ)^2$<\/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\"><strong>Step 2: Plug values into the formula<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>13.5 = \\frac{v_0^2 (\\sin 15^\\circ)^2}{2 \\cdot 9.8}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\sin 15^\\circ \\approx 0.2588 \\quad \\Rightarrow \\quad (\\sin 15^\\circ)^2 \\approx 0.067<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>13.5 = \\frac{v_0^2 \\cdot 0.067}{19.6}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Multiply both sides by 19.6:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>13.5 \\cdot 19.6 = v_0^2 \\cdot 0.067<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>264.6 = v_0^2 \\cdot 0.067<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now divide both sides by 0.067:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>v_0^2 = \\frac{264.6}{0.067} \\approx 3,948.5<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>v_0 = \\sqrt{3948.5} \\approx \\boxed{62.8 \\, \\text{m\/s}}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (300+ words)<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">To solve this problem, we focus on the <strong>vertical motion<\/strong> of the projectile. When a projectile is launched, its motion can be split into horizontal and vertical components. The maximum height that a projectile reaches is influenced only by its vertical velocity component and gravity.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The vertical component of the initial velocity is given by $v_0 \\sin \\theta$, and using basic kinematics, the formula for the <strong>maximum vertical height<\/strong> (assuming upward motion until vertical velocity becomes zero) is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>h = \\frac{(v_0 \\sin \\theta)^2}{2g}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In this case, we know the projectile reaches a <strong>maximum height of 13.5 m<\/strong>, and it was launched at <strong>15.0\u00b0<\/strong>, a relatively shallow angle. That means the vertical component of its velocity is relatively small, and it needs a large total launch speed to achieve that height.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">By solving for $v_0$, we rearranged the formula and substituted known values: $g = 9.8 \\, \\text{m\/s}^2$ and $\\sin(15^\\circ) \\approx 0.2588$. Squaring the sine value, multiplying and isolating $v_0$, we find the launch speed to be <strong>approximately 62.8 m\/s<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This result shows that even at a low angle, a high speed can allow the projectile to reach a significant height. Understanding how vertical and horizontal components affect projectile motion is key in solving such physics problems.<\/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 To determine the launch speed of a projectile that clears a barrier of 13.5 m at a launch angle [&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-213247","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213247","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=213247"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213247\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=213247"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=213247"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=213247"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}