{"id":195457,"date":"2025-02-28T09:25:54","date_gmt":"2025-02-28T09:25:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=195457"},"modified":"2025-02-28T09:25:57","modified_gmt":"2025-02-28T09:25:57","slug":"a-ball-is-to-be-shot-from-level-ground-toward-a-wall-at-distance-x","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/28\/a-ball-is-to-be-shot-from-level-ground-toward-a-wall-at-distance-x\/","title":{"rendered":"A ball is to be shot from level ground toward a wall at distance x"},"content":{"rendered":"\n<p>A ball is to be shot from level ground toward a wall at distance x (Figure a). (Figure b) shows the y component vy of the ball&#8217;s velocity just as it would reach the wall, as a function of that distance x. The scaling is set by vys= 5.0 m\/s and xs = 20 m.What is the launch angle?<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-1641.png\" alt=\"\" class=\"wp-image-195458\"\/><\/figure>\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 determine the launch angle (\u03b8) of a projectile based on the given graph of the vertical velocity component (vy) as a function of horizontal distance (x), we can analyze the relationship between these variables in projectile motion.<\/p>\n\n\n\n<p><strong>Understanding Projectile Motion:<\/strong><\/p>\n\n\n\n<p>In projectile motion, the horizontal (vx) and vertical (vy) components of velocity are influenced differently:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Horizontal Component (vx):<\/strong> Remains constant throughout the flight, as there is no acceleration in the horizontal direction (assuming air resistance is negligible).<\/li>\n\n\n\n<li><strong>Vertical Component (vy):<\/strong> Changes linearly over time due to the constant acceleration caused by gravity (g \u2248 9.8 m\/s\u00b2 downward).<\/li>\n<\/ul>\n\n\n\n<p>The vertical velocity at any point can be described by:<\/p>\n\n\n\n<p>vy = v\u2080y &#8211; g * t<\/p>\n\n\n\n<p>where v\u2080y is the initial vertical velocity component, and t is the time elapsed.<\/p>\n\n\n\n<p><strong>Analyzing the Graph:<\/strong><\/p>\n\n\n\n<p>The provided graph shows vy as a function of x, with scaling factors vys = 5.0 m\/s and xs = 20 m. This indicates that at x = 20 m, vy = 5.0 m\/s.<\/p>\n\n\n\n<p><strong>Relating Horizontal Distance and Time:<\/strong><\/p>\n\n\n\n<p>The horizontal distance traveled is given by:<\/p>\n\n\n\n<p>x = vx * t<\/p>\n\n\n\n<p>Solving for time (t):<\/p>\n\n\n\n<p>t = x \/ vx<\/p>\n\n\n\n<p><strong>Expressing vy in Terms of x:<\/strong><\/p>\n\n\n\n<p>Substitute t into the equation for vy:<\/p>\n\n\n\n<p>vy = v\u2080y &#8211; g * (x \/ vx)<\/p>\n\n\n\n<p>Since vx = v\u2080 * cos(\u03b8) and v\u2080y = v\u2080 * sin(\u03b8), we have:<\/p>\n\n\n\n<p>vy = v\u2080 * sin(\u03b8) &#8211; g * (x \/ (v\u2080 * cos(\u03b8)))<\/p>\n\n\n\n<p>Simplifying:<\/p>\n\n\n\n<p>vy = v\u2080 * sin(\u03b8) &#8211; (g * x) \/ (v\u2080 * cos(\u03b8))<\/p>\n\n\n\n<p><strong>Using the Given Data:<\/strong><\/p>\n\n\n\n<p>At x = 20 m, vy = 5.0 m\/s. Plugging these values into the equation:<\/p>\n\n\n\n<p>5.0 m\/s = v\u2080 * sin(\u03b8) &#8211; (9.8 m\/s\u00b2 * 20 m) \/ (v\u2080 * cos(\u03b8))<\/p>\n\n\n\n<p>Rearranging terms:<\/p>\n\n\n\n<p>5.0 m\/s = v\u2080 * sin(\u03b8) &#8211; (196 m\u00b2\/s\u00b2) \/ (v\u2080 * cos(\u03b8))<\/p>\n\n\n\n<p>Let k = v\u2080 * cos(\u03b8). Then, v\u2080 * sin(\u03b8) = k * tan(\u03b8), and the equation becomes:<\/p>\n\n\n\n<p>5.0 m\/s = k * tan(\u03b8) &#8211; 196 m\u00b2\/s\u00b2 \/ k<\/p>\n\n\n\n<p>Multiplying through by k to clear the denominator:<\/p>\n\n\n\n<p>5.0 m\/s * k = k\u00b2 * tan(\u03b8) &#8211; 196 m\u00b2\/s\u00b2<\/p>\n\n\n\n<p>Rearranging:<\/p>\n\n\n\n<p>k\u00b2 * tan(\u03b8) &#8211; 5.0 m\/s * k &#8211; 196 m\u00b2\/s\u00b2 = 0<\/p>\n\n\n\n<p>This is a quadratic equation in terms of k. Solving for k using the quadratic formula:<\/p>\n\n\n\n<p>k = [5.0 m\/s \u00b1 sqrt((5.0 m\/s)\u00b2 + 4 * tan(\u03b8) * 196 m\u00b2\/s\u00b2)] \/ (2 * tan(\u03b8))<\/p>\n\n\n\n<p>Since k = v\u2080 * cos(\u03b8), we can express v\u2080 as:<\/p>\n\n\n\n<p>v\u2080 = k \/ cos(\u03b8)<\/p>\n\n\n\n<p><strong>Determining the Launch Angle (\u03b8):<\/strong><\/p>\n\n\n\n<p>To find \u03b8, we need to know v\u2080 or have additional information. However, if we assume that the projectile reaches its maximum height at the wall (x = 20 m), then vy = 0 m\/s at that point. This simplifies our earlier equation:<\/p>\n\n\n\n<p>0 = v\u2080 * sin(\u03b8) &#8211; (9.8 m\/s\u00b2 * 20 m) \/ (v\u2080 * cos(\u03b8))<\/p>\n\n\n\n<p>Solving for \u03b8:<\/p>\n\n\n\n<p>v\u2080 * sin(\u03b8) = (9.8 m\/s\u00b2 * 20 m) \/ (v\u2080 * cos(\u03b8))<\/p>\n\n\n\n<p>tan(\u03b8) = (9.8 m\/s\u00b2 * 20 m) \/ v\u2080\u00b2<\/p>\n\n\n\n<p>\u03b8 = arctan((9.8 m\/s\u00b2 * 20 m) \/ v\u2080\u00b2)<\/p>\n\n\n\n<p>Without the exact value of v\u2080, we cannot determine a numerical value for \u03b8. Additional information, such as the initial speed or the time of flight, would be necessary to calculate the precise launch angle.<\/p>\n\n\n\n<p>In summary, the launch angle \u03b8 can be determined from the relationship between vy and x, but specific numerical values require knowledge of the initial velocity or other parameters of the projectile&#8217;s motion.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A ball is to be shot from level ground toward a wall at distance x (Figure a). (Figure b) shows the y component vy of the ball&#8217;s velocity just as it would reach the wall, as a function of that distance x. The scaling is set by vys= 5.0 m\/s and xs = 20 m.What [&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-195457","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195457","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=195457"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195457\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195457"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195457"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195457"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}