{"id":242497,"date":"2025-07-04T08:03:33","date_gmt":"2025-07-04T08:03:33","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=242497"},"modified":"2025-07-04T08:03:35","modified_gmt":"2025-07-04T08:03:35","slug":"a-basketball-player-is-standing-on-the-floor-10-0-m-from-the-basket-as-in-figure-p4-60","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/04\/a-basketball-player-is-standing-on-the-floor-10-0-m-from-the-basket-as-in-figure-p4-60\/","title":{"rendered":"A basketball player is standing on the floor\u00a010.0\u00a0m\u00a0from the basket as in Figure P4.60."},"content":{"rendered":"\n
A basketball player is standing on the floor 10.0 m from the basket as in Figure P4.60.<\/pre>\n\n\n\n
\"\"<\/figure>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Here are the answers to the physics problem.<\/p>\n\n\n\n
(a) 9.80 m\/s\u00b2 downwards<\/strong>
(b) 10.7 m\/s<\/strong><\/p>\n\n\n\n
Explanation<\/h3>\n\n\n\n
(a) Acceleration at the Highest Point<\/strong><\/p>\n\n\n\n
Once the basketball leaves the player’s hands, the only significant force acting on it is gravity (assuming air resistance is negligible). According to Newton’s second law, this constant downward force produces a constant downward acceleration. This acceleration is the acceleration due to gravity, g, which is approximately 9.80 m\/s\u00b2.<\/p>\n\n\n\n
This acceleration remains constant throughout the entire flight of the ball. Even at the highest point of its trajectory, where the vertical component of its velocity is momentarily zero, the force of gravity is still acting on it. Therefore, the acceleration of the basketball at the highest point is 9.80 m\/s\u00b2 directed downwards.<\/p>\n\n\n\n
(b) Initial Speed of the Throw<\/strong><\/p>\n\n\n\n
To find the initial speed (v\u2080) required for the ball to reach the hoop, we analyze its motion in two dimensions: horizontal and vertical.<\/p>\n\n\n\n
    \n
  1. Set up the coordinate system:<\/strong>\u00a0Let the initial position of the ball be (x\u2080, y\u2080) = (0, 2.00 m). The final position, the hoop, is at (x, y) = (10.0 m, 3.05 m). The launch angle \u03b8 is 40.0\u00b0.<\/li>\n\n\n\n
  2. Analyze the motion:<\/strong>\n
      \n
    • Horizontal Motion:<\/strong>\u00a0The horizontal velocity (v\u2093) is constant.
      x = v\u2080\u2093 * t
      v\u2080\u2093 = v\u2080 * cos(40.0\u00b0)
      So, 10.0 = (v\u2080 * cos(40.0\u00b0)) * t
      We can express the time of flight (t) as: t = 10.0 \/ (v\u2080 * cos(40.0\u00b0))<\/li>\n\n\n\n
    • Vertical Motion:<\/strong>\u00a0The vertical motion has a constant downward acceleration of g = 9.80 m\/s\u00b2.
      y = y\u2080 + v\u2080y * t – (1\/2)gt\u00b2
      v\u2080y = v\u2080 * sin(40.0\u00b0)
      Substituting the known values:
      3.05 = 2.00 + (v\u2080 * sin(40.0\u00b0)) * t – (1\/2)(9.80)t\u00b2<\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • Solve for the initial speed (v\u2080):<\/strong>
      Now, substitute the expression for t from the horizontal motion into the vertical motion equation. This eliminates t and allows us to solve for v\u2080.1.05 = (v\u2080 * sin(40.0\u00b0)) * [10.0 \/ (v\u2080 * cos(40.0\u00b0))] – 4.90 * [10.0 \/ (v\u2080 * cos(40.0\u00b0))]\u00b2
      1.05 = 10.0 * tan(40.0\u00b0) – 490 \/ (v\u2080\u00b2 * cos\u00b2(40.0\u00b0))Now, solve for v\u2080:
      1.05 = 10.0 * (0.8391) – 490 \/ (v\u2080\u00b2 * (0.7660)\u00b2)
      1.05 = 8.391 – 490 \/ (v\u2080\u00b2 * 0.5868)
      1.05 = 8.391 – 835.0 \/ v\u2080\u00b2
      835.0 \/ v\u2080\u00b2 = 8.391 – 1.05 = 7.341
      v\u2080\u00b2 = 835.0 \/ 7.341 = 113.75
      v\u2080 = \u221a113.75 \u2248 10.665 m\/sRounding to three significant figures, the required initial speed is\u00a010.7 m\/s<\/strong>.<\/li>\n<\/ol>\n\n\n\n
      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
      A basketball player is standing on the floor 10.0 m from the basket as in Figure P4.60. The Correct Answer and Explanation is: Here are the answers to the physics problem. (a) 9.80 m\/s\u00b2 downwards(b) 10.7 m\/s Explanation (a) Acceleration at the Highest Point Once the basketball leaves the player’s hands, the only significant force acting on it […]<\/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-242497","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/242497","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=242497"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/242497\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=242497"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=242497"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=242497"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}