{"id":190605,"date":"2025-02-12T17:24:31","date_gmt":"2025-02-12T17:24:31","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=190605"},"modified":"2025-02-12T17:24:34","modified_gmt":"2025-02-12T17:24:34","slug":"a-basketball-player-is-standing-on-the-floor-10-0-m-from-the-basket-as-in-the-figure-below","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/12\/a-basketball-player-is-standing-on-the-floor-10-0-m-from-the-basket-as-in-the-figure-below\/","title":{"rendered":"A basketball player is standing on the floor 10.0 m from the basket as in the figure below"},"content":{"rendered":"\n

A basketball player is standing on the floor 10.0 m from the basket as in the figure below. The height of the basket is H = 3.05 m, and he shoots the ball at an angle, ? = 42.0A????1, with the horizontal from a height of h = 2.20 m.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(a) What is the acceleration of the basketball at the highest point in its trajectory?<\/p>\n\n\n\n

magnitude
direction —Select— upward, or downward<\/p>\n\n\n\n

(b) At what speed must the player throw the basketball so that the ball goes through the hoop without striking the backboard?
m\/s<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

(a) What is the acceleration of the basketball at the highest point in its trajectory?<\/h3>\n\n\n\n

At the highest point in its trajectory, the basketball has no vertical velocity. However, the vertical acceleration due to gravity is still acting on it, pulling it downward.<\/p>\n\n\n\n

    \n
  • Magnitude<\/strong>: The magnitude of the acceleration is constant throughout the entire flight of the ball and is equal to the acceleration due to gravity. This value is approximately ( 9.81 \\, \\text{m\/s}^2 ).<\/li>\n\n\n\n
  • Direction<\/strong>: The direction of the acceleration is downward since gravity pulls the ball toward the ground.<\/li>\n<\/ul>\n\n\n\n

    Answer for (a):<\/strong><\/p>\n\n\n\n

      \n
    • Magnitude: ( 9.81 \\, \\text{m\/s}^2 )<\/li>\n\n\n\n
    • Direction: Downward<\/li>\n<\/ul>\n\n\n\n

      (b) At what speed must the player throw the basketball so that the ball goes through the hoop without striking the backboard?<\/h3>\n\n\n\n

      To solve this, we will use projectile motion equations. The ball’s initial velocity can be split into two components:<\/p>\n\n\n\n

        \n
      • The horizontal component (( v_{0x} )) and<\/li>\n\n\n\n
      • The vertical component (( v_{0y} )).<\/li>\n<\/ul>\n\n\n\n

        Given:<\/p>\n\n\n\n

          \n
        • Horizontal distance (range): ( x = 10.0 \\, \\text{m} )<\/li>\n\n\n\n
        • Initial vertical height: ( y_0 = 2.20 \\, \\text{m} )<\/li>\n\n\n\n
        • Height of the basket: ( y_{\\text{basket}} = 3.05 \\, \\text{m} )<\/li>\n\n\n\n
        • Launch angle: ( \\theta = 42^\\circ )<\/li>\n\n\n\n
        • Gravity: ( g = 9.81 \\, \\text{m\/s}^2 )<\/li>\n<\/ul>\n\n\n\n

          We know that the horizontal velocity is constant throughout the flight because there is no horizontal acceleration. The equations for projectile motion are:<\/p>\n\n\n\n

            \n
          1. Horizontal Motion<\/strong>:
            [
            x = v_0 \\cdot \\cos(\\theta) \\cdot t
            ]<\/li>\n\n\n\n
          2. Vertical Motion<\/strong>:
            [
            y = y_0 + v_0 \\cdot \\sin(\\theta) \\cdot t – \\frac{1}{2} g t^2
            ]<\/li>\n<\/ol>\n\n\n\n

            To find the velocity ( v_0 ), we need to eliminate time (( t )) from the equations. First, solve the horizontal motion equation for time (( t )):<\/p>\n\n\n\n

            [
            t = \\frac{x}{v_0 \\cos(\\theta)}
            ]<\/p>\n\n\n\n

            Substitute this expression for ( t ) into the vertical motion equation:<\/p>\n\n\n\n

            [
            y_{\\text{basket}} = y_0 + v_0 \\cdot \\sin(\\theta) \\cdot \\frac{x}{v_0 \\cdot \\cos(\\theta)} – \\frac{1}{2} g \\left( \\frac{x}{v_0 \\cdot \\cos(\\theta)} \\right)^2
            ]<\/p>\n\n\n\n

            Simplify this equation and solve for ( v_0 ). Using this approach, we can find the required initial velocity to make the basketball go through the hoop.<\/p>\n\n\n\n

            Let’s calculate this:<\/p>\n\n\n\n

            [
            y_{\\text{basket}} = y_0 + x \\cdot \\tan(\\theta) – \\frac{g x^2}{2 v_0^2 \\cdot \\cos^2(\\theta)}
            ]<\/p>\n\n\n\n

            Substitute the known values and solve for ( v_0 ):<\/p>\n\n\n\n

            [
            3.05 = 2.20 + 10.0 \\cdot \\tan(42^\\circ) – \\frac{9.81 \\cdot 10.0^2}{2 v_0^2 \\cdot \\cos^2(42^\\circ)}
            ]<\/p>\n\n\n\n

            I’ll calculate this now.<\/p>\n\n\n\n

            The required initial velocity for the basketball to go through the hoop without striking the backboard is approximately ( 10.44 \\, \\text{m\/s} ).<\/p>\n\n\n\n

            Explanation:<\/h3>\n\n\n\n

            We derived the equation based on the projectile motion principles and then solved for the initial velocity ( v_0 ) that would allow the ball to travel a horizontal distance of 10.0 meters and reach the height of 3.05 meters at the basket.<\/p>\n\n\n\n

            The negative value for velocity is discarded, as velocity cannot be negative in this context. Therefore, the player must throw the ball with a speed of around ( 10.44 \\, \\text{m\/s} ) at an angle of ( 42^\\circ ) to ensure it reaches the basket without hitting the backboard.<\/p>\n","protected":false},"excerpt":{"rendered":"

            A basketball player is standing on the floor 10.0 m from the basket as in the figure below. The height of the basket is H = 3.05 m, and he shoots the ball at an angle, ? = 42.0A????1, with the horizontal from a height of h = 2.20 m. (a) What is the acceleration […]<\/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-190605","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190605","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=190605"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190605\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190605"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190605"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190605"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}