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
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the launch speed<\/strong> of a projectile that reaches a maximum height of 13.5 m<\/strong> when launched at an angle of 15.0\u00b0<\/strong>, we need to apply principles of projectile motion<\/strong> and kinematics<\/strong>.<\/p>\n\n\n\n The projectile\u2019s launch speed<\/strong> is approximately 37.9 m\/s<\/strong>.<\/p>\n\n\n\n A projectile reaches its maximum height<\/strong> when its vertical velocity component becomes zero<\/strong> (i.e., at the peak of the trajectory). We can use kinematic equations to analyze this vertical motion.<\/p>\n\n\n\n Let\u2019s define the known variables:<\/p>\n\n\n\n Let ( v_0 ) be the initial launch speed<\/strong>, and ( v_{0y} ) be the initial vertical velocity component<\/strong>, which is given by:<\/p>\n\n\n\n [ We use the following kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:<\/p>\n\n\n\n [ Since ( v_y = 0 ) at the peak:<\/p>\n\n\n\n [ Substitute the known values:<\/p>\n\n\n\n [ Now recall that:<\/p>\n\n\n\n [ Using ( \\sin(15^\\circ) \\approx 0.2588 ):<\/p>\n\n\n\n [ \ud83d\udd01 Correction<\/strong>: There was an error in approximating earlier. Let’s redo the calculation:<\/p>\n\n\n\n [ [ 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 determine the launch speed of a projectile that reaches a maximum height of 13.5 m when […]<\/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-209727","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209727","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=209727"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209727\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=209727"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=209727"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=209727"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n\ud83d\udd39 Final Answer:<\/strong><\/h3>\n\n\n\n
\n\n\n\n\ud83d\udd39 Step-by-Step Explanation (300+ words):<\/strong><\/h3>\n\n\n\n
\n
v_{0y} = v_0 \\cdot \\sin(\\theta)
]<\/p>\n\n\n\n
v_y^2 = v_{0y}^2 – 2g h
]<\/p>\n\n\n\n
0 = v_{0y}^2 – 2gh
\\Rightarrow v_{0y}^2 = 2gh
\\Rightarrow v_{0y} = \\sqrt{2gh}
]<\/p>\n\n\n\n
v_{0y} = \\sqrt{2 \\cdot 9.8 \\cdot 13.5}
= \\sqrt{264.6}
\\approx 16.27 \\, \\text{m\/s}
]<\/p>\n\n\n\n
v_{0y} = v_0 \\cdot \\sin(15^\\circ)
\\Rightarrow v_0 = \\frac{v_{0y}}{\\sin(15^\\circ)}
]<\/p>\n\n\n\n
v_0 = \\frac{16.27}{0.2588} \\approx 62.9 \\, \\text{m\/s}
]<\/p>\n\n\n\n
v_{0y} = \\sqrt{2 \\cdot 9.8 \\cdot 13.5} = \\sqrt{264.6} \\approx 16.27 \\, \\text{m\/s}
]<\/p>\n\n\n\n
v_0 = \\frac{16.27}{\\sin(15^\\circ)} = \\frac{16.27}{0.2588} \\approx \\boxed{62.9 \\, \\text{m\/s}}
]<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer: 62.9 m\/s<\/strong><\/h3>\n","protected":false},"excerpt":{"rendered":"