{"id":186727,"date":"2025-01-26T06:37:35","date_gmt":"2025-01-26T06:37:35","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=186727"},"modified":"2025-01-26T06:37:42","modified_gmt":"2025-01-26T06:37:42","slug":"a-projectile-is-launched-from-the-ground-with-an-initial-velocity-of-30-m-s-at-an-angle-of-60-degrees-above-the-horizontal","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/26\/a-projectile-is-launched-from-the-ground-with-an-initial-velocity-of-30-m-s-at-an-angle-of-60-degrees-above-the-horizontal\/","title":{"rendered":"A projectile is launched from the ground with an initial velocity of 30 m\/s at an angle of 60 degrees above the horizontal"},"content":{"rendered":"\n<p>A projectile is launched from the ground with an initial velocity of 30 m\/s at an angle of 60 degrees above the horizontal. Determine the maximum height reached by the projectile and the total time of flight. Assume no air resistance and take g = 9.8 m\/s\u00b2.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To solve this problem, we will break the projectile\u2019s motion into two components: horizontal and vertical. We will use the equations of motion for projectile trajectories.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initial velocity ( v_0 = 30 \\, \\text{m\/s} )<\/li>\n\n\n\n<li>Launch angle ( \\theta = 60^\\circ )<\/li>\n\n\n\n<li>Gravitational acceleration ( g = 9.8 \\, \\text{m\/s}^2 )<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Break the initial velocity into components<\/h3>\n\n\n\n<p>We can decompose the initial velocity into horizontal (( v_{0x} )) and vertical (( v_{0y} )) components:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( v_{0x} = v_0 \\cdot \\cos(\\theta) )<\/li>\n\n\n\n<li>( v_{0y} = v_0 \\cdot \\sin(\\theta) )<\/li>\n<\/ul>\n\n\n\n<p>So:<br>[<br>v_{0x} = 30 \\cdot \\cos(60^\\circ) = 30 \\cdot 0.5 = 15 \\, \\text{m\/s}<br>]<br>[<br>v_{0y} = 30 \\cdot \\sin(60^\\circ) = 30 \\cdot \\frac{\\sqrt{3}}{2} = 30 \\cdot 0.866 = 25.98 \\, \\text{m\/s}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Time to reach the maximum height<\/h3>\n\n\n\n<p>At the maximum height, the vertical velocity becomes zero (( v_y = 0 )). Using the kinematic equation:<br>[<br>v_y = v_{0y} &#8211; g \\cdot t<br>]<br>Setting ( v_y = 0 ) and solving for time to reach the maximum height:<br>[<br>0 = 25.98 &#8211; 9.8 \\cdot t<br>]<br>[<br>t = \\frac{25.98}{9.8} = 2.65 \\, \\text{seconds}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Maximum height<\/h3>\n\n\n\n<p>The maximum height ( H ) can be found using the following kinematic equation:<br>[<br>H = v_{0y} \\cdot t &#8211; \\frac{1}{2} g \\cdot t^2<br>]<br>Substituting ( v_{0y} = 25.98 \\, \\text{m\/s} ) and ( t = 2.65 \\, \\text{seconds} ):<br>[<br>H = 25.98 \\cdot 2.65 &#8211; \\frac{1}{2} \\cdot 9.8 \\cdot (2.65)^2<br>]<br>[<br>H = 68.88 &#8211; 34.06 = 34.82 \\, \\text{m}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Total time of flight<\/h3>\n\n\n\n<p>The total time of flight is twice the time to reach the maximum height, since the time taken to rise is equal to the time taken to fall:<br>[<br>T = 2 \\cdot 2.65 = 5.30 \\, \\text{seconds}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Maximum height ( H = 34.82 \\, \\text{m} )<\/li>\n\n\n\n<li>Total time of flight ( T = 5.30 \\, \\text{seconds} )<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The projectile is launched at an angle, so its motion is influenced by both vertical and horizontal components. The vertical component is responsible for the rise and fall of the projectile, while the horizontal component determines the range. By breaking the initial velocity into components and using the kinematic equations, we calculated the maximum height the projectile reaches and the total time it remains in the air.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A projectile is launched from the ground with an initial velocity of 30 m\/s at an angle of 60 degrees above the horizontal. Determine the maximum height reached by the projectile and the total time of flight. Assume no air resistance and take g = 9.8 m\/s\u00b2. The Correct Answer and Explanation is : To [&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-186727","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186727","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=186727"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186727\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186727"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186727"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186727"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}