{"id":217806,"date":"2025-05-22T13:54:59","date_gmt":"2025-05-22T13:54:59","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=217806"},"modified":"2025-05-22T13:55:01","modified_gmt":"2025-05-22T13:55:01","slug":"the-highest-barrier-that-a-projectile-can-clear-is-13-5-m-5","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/22\/the-highest-barrier-that-a-projectile-can-clear-is-13-5-m-5\/","title":{"rendered":"The highest barrier that a projectile can clear is 13.5 m"},"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 clears a maximum height (vertical displacement) of <strong>13.5 m<\/strong> when launched at an angle of <strong>15.0\u00b0<\/strong> above the horizontal, we can use <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 to find vertical velocity<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The vertical component of the initial velocity can be related to the maximum height using the following kinematic equation:<\/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, the vertical velocity $v_y$ = 0 (because the projectile momentarily stops moving upward), so:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>0 = v_{y0}^2 &#8211; 2gh<br>\\Rightarrow v_{y0}^2 = 2gh<br>\\Rightarrow v_{y0} = \\sqrt{2gh}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Plug in:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$g = 9.8 \\, \\text{m\/s}^2$<\/li>\n\n\n\n<li>$h = 13.5 \\, \\text{m}$<\/li>\n<\/ul>\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<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Use angle to find total launch speed<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Now we use the angle to relate vertical velocity to total velocity.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>v_{y0} = v_0 \\sin \\theta<br>\\Rightarrow v_0 = \\frac{v_{y0}}{\\sin \\theta}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\theta = 15^\\circ \\Rightarrow \\sin 15^\\circ \\approx 0.2588<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>v_0 = \\frac{16.27}{0.2588} \\approx 62.87 \\, \\text{m\/s}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\boxed{62.9 \\, \\text{m\/s}}<br>$$<\/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\">In projectile motion, the <strong>maximum height<\/strong> a projectile reaches is determined solely by the <strong>vertical component<\/strong> of its velocity, since only vertical motion is influenced by gravity. At the top of its arc, the projectile&#8217;s vertical velocity becomes zero \u2014 it stops rising and begins to fall back down. This is a key point in solving this problem.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">We\u2019re told the projectile reaches a maximum vertical height of <strong>13.5 m<\/strong>, and it was launched at an angle of <strong>15\u00b0<\/strong>. Since only the <strong>vertical part<\/strong> of the launch speed contributes to reaching this height, we start by using the kinematic equation that relates initial vertical velocity ($v_{y0}$) to the height. The equation $v_y^2 = v_{y0}^2 &#8211; 2gh$ simplifies to $v_{y0}^2 = 2gh$ when $v_y = 0$ at the peak.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">After calculating the vertical component of the initial velocity to be about <strong>16.27 m\/s<\/strong>, we then use trigonometry to relate this vertical component to the full launch velocity. The sine of the launch angle gives us this relationship: $v_{y0} = v_0 \\sin \\theta$. By isolating $v_0$, we can solve for the full initial speed.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Using $\\sin(15\u00b0) \\approx 0.2588$, the total launch speed comes out to approximately <strong>62.9 m\/s<\/strong>. This is the speed required to launch the projectile at a 15\u00b0 angle and still clear a vertical height of 13.5 m.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This problem is a classic example of how <strong>kinematics<\/strong> and <strong>trigonometry<\/strong> combine in physics to analyze motion in two dimensions.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-74.png\" alt=\"\" class=\"wp-image-217807\"\/><\/figure>\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 clears a maximum height (vertical displacement) of 13.5 m [&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-217806","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217806","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=217806"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217806\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217806"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217806"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217806"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}