{"id":88817,"date":"2025-06-01T07:21:47","date_gmt":"2025-06-01T07:21:47","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88817"},"modified":"2025-06-01T07:21:47","modified_gmt":"2025-06-01T07:21:47","slug":"a-tennis-ball-is-thrown-in-the-vertically-upward-direction-and-the-bal","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-tennis-ball-is-thrown-in-the-vertically-upward-direction-and-the-bal\/","title":{"rendered":"A tennis ball is thrown in the vertically upward direction and the bal"},"content":{"rendered":"

A tennis ball is thrown in the vertically upward direction and the ball attains a maximum height of 20 m. The ball was thrown approximately with an upward velocity of<\/p>\n

[amp_mcq option1=”8 m\/s” option2=”12 m\/s” option3=”16 m\/s” option4=”20 m\/s” correct=”option4″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-2 – 2021<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct option is D. Using the kinematic equation $v^2 = u^2 + 2as$, where $v$ is the final velocity (0 m\/s at max height), $u$ is the initial velocity, $a$ is the acceleration due to gravity (-g), and $s$ is the height (20 m).
\n<\/section>\n
\nAt the maximum height of a vertically thrown object, its instantaneous velocity is zero. The acceleration acting on the object throughout its flight (ignoring air resistance) is the constant acceleration due to gravity, acting downwards. Taking the upward direction as positive, $a = -g$. Using $g \\approx 10 \\, \\text{m\/s}^2$ for approximation: $0^2 = u^2 + 2(-10)(20) \\implies 0 = u^2 – 400 \\implies u^2 = 400 \\implies u = 20 \\, \\text{m\/s}$. Using $g \\approx 9.8 \\, \\text{m\/s}^2$: $0^2 = u^2 + 2(-9.8)(20) \\implies 0 = u^2 – 392 \\implies u^2 = 392 \\implies u \\approx 19.8 \\, \\text{m\/s}$. Both approximations are closest to 20 m\/s among the given options.
\n<\/section>\n
\nThis is a standard problem in projectile motion under constant acceleration (gravity). The time taken to reach the maximum height can be found using $v = u + at$. The total time of flight is twice the time to reach the maximum height (assuming it lands at the same level).
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A tennis ball is thrown in the vertically upward direction and the ball attains a maximum height of 20 m. The ball was thrown approximately with an upward velocity of [amp_mcq option1=”8 m\/s” option2=”12 m\/s” option3=”16 m\/s” option4=”20 m\/s” correct=”option4″] This question was previously asked in UPSC NDA-2 – 2021 Download PDFAttempt Online The correct … <\/p>\n

Detailed SolutionA tennis ball is thrown in the vertically upward direction and the bal<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1094],"tags":[1110,1421,1128],"class_list":["post-88817","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1110","tag-motion-under-gravity","tag-physics","no-featured-image-padding"],"yoast_head":"\nA tennis ball is thrown in the vertically upward direction and the bal<\/title>\n<meta name=\"description\" content=\"The correct option is D. Using the kinematic equation $v^2 = u^2 + 2as$, where $v$ is the final velocity (0 m\/s at max height), $u$ is the initial velocity, $a$ is the acceleration due to gravity (-g), and $s$ is the height (20 m). At the maximum height of a vertically thrown object, its instantaneous velocity is zero. The acceleration acting on the object throughout its flight (ignoring air resistance) is the constant acceleration due to gravity, acting downwards. Taking the upward direction as positive, $a = -g$. Using $g approx 10 , text{m\/s}^2$ for approximation: $0^2 = u^2 + 2(-10)(20) implies 0 = u^2 - 400 implies u^2 = 400 implies u = 20 , text{m\/s}$. Using $g approx 9.8 , text{m\/s}^2$: $0^2 = u^2 + 2(-9.8)(20) implies 0 = u^2 - 392 implies u^2 = 392 implies u approx 19.8 , text{m\/s}$. Both approximations are closest to 20 m\/s among the given options.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/a-tennis-ball-is-thrown-in-the-vertically-upward-direction-and-the-bal\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A tennis ball is thrown in the vertically upward direction and the bal\" \/>\n<meta property=\"og:description\" content=\"The correct option is D. Using the kinematic equation $v^2 = u^2 + 2as$, where $v$ is the final velocity (0 m\/s at max height), $u$ is the initial velocity, $a$ is the acceleration due to gravity (-g), and $s$ is the height (20 m). At the maximum height of a vertically thrown object, its instantaneous velocity is zero. The acceleration acting on the object throughout its flight (ignoring air resistance) is the constant acceleration due to gravity, acting downwards. Taking the upward direction as positive, $a = -g$. Using $g approx 10 , text{m\/s}^2$ for approximation: $0^2 = u^2 + 2(-10)(20) implies 0 = u^2 - 400 implies u^2 = 400 implies u = 20 , text{m\/s}$. Using $g approx 9.8 , text{m\/s}^2$: $0^2 = u^2 + 2(-9.8)(20) implies 0 = u^2 - 392 implies u^2 = 392 implies u approx 19.8 , text{m\/s}$. Both approximations are closest to 20 m\/s among the given options.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-tennis-ball-is-thrown-in-the-vertically-upward-direction-and-the-bal\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:21:47+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"A tennis ball is thrown in the vertically upward direction and the bal","description":"The correct option is D. Using the kinematic equation $v^2 = u^2 + 2as$, where $v$ is the final velocity (0 m\/s at max height), $u$ is the initial velocity, $a$ is the acceleration due to gravity (-g), and $s$ is the height (20 m). At the maximum height of a vertically thrown object, its instantaneous velocity is zero. The acceleration acting on the object throughout its flight (ignoring air resistance) is the constant acceleration due to gravity, acting downwards. Taking the upward direction as positive, $a = -g$. Using $g approx 10 , text{m\/s}^2$ for approximation: $0^2 = u^2 + 2(-10)(20) implies 0 = u^2 - 400 implies u^2 = 400 implies u = 20 , text{m\/s}$. Using $g approx 9.8 , text{m\/s}^2$: $0^2 = u^2 + 2(-9.8)(20) implies 0 = u^2 - 392 implies u^2 = 392 implies u approx 19.8 , text{m\/s}$. Both approximations are closest to 20 m\/s among the given options.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/a-tennis-ball-is-thrown-in-the-vertically-upward-direction-and-the-bal\/","og_locale":"en_US","og_type":"article","og_title":"A tennis ball is thrown in the vertically upward direction and the bal","og_description":"The correct option is D. Using the kinematic equation $v^2 = u^2 + 2as$, where $v$ is the final velocity (0 m\/s at max height), $u$ is the initial velocity, $a$ is the acceleration due to gravity (-g), and $s$ is the height (20 m). At the maximum height of a vertically thrown object, its instantaneous velocity is zero. The acceleration acting on the object throughout its flight (ignoring air resistance) is the constant acceleration due to gravity, acting downwards. Taking the upward direction as positive, $a = -g$. Using $g approx 10 , text{m\/s}^2$ for approximation: $0^2 = u^2 + 2(-10)(20) implies 0 = u^2 - 400 implies u^2 = 400 implies u = 20 , text{m\/s}$. Using $g approx 9.8 , text{m\/s}^2$: $0^2 = u^2 + 2(-9.8)(20) implies 0 = u^2 - 392 implies u^2 = 392 implies u approx 19.8 , text{m\/s}$. Both approximations are closest to 20 m\/s among the given options.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-tennis-ball-is-thrown-in-the-vertically-upward-direction-and-the-bal\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:21:47+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-tennis-ball-is-thrown-in-the-vertically-upward-direction-and-the-bal\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-tennis-ball-is-thrown-in-the-vertically-upward-direction-and-the-bal\/","name":"A tennis ball is thrown in the vertically upward direction and the bal","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:21:47+00:00","dateModified":"2025-06-01T07:21:47+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct option is D. Using the kinematic equation $v^2 = u^2 + 2as$, where $v$ is the final velocity (0 m\/s at max height), $u$ is the initial velocity, $a$ is the acceleration due to gravity (-g), and $s$ is the height (20 m). At the maximum height of a vertically thrown object, its instantaneous velocity is zero. The acceleration acting on the object throughout its flight (ignoring air resistance) is the constant acceleration due to gravity, acting downwards. Taking the upward direction as positive, $a = -g$. Using $g \\approx 10 \\, \\text{m\/s}^2$ for approximation: $0^2 = u^2 + 2(-10)(20) \\implies 0 = u^2 - 400 \\implies u^2 = 400 \\implies u = 20 \\, \\text{m\/s}$. Using $g \\approx 9.8 \\, \\text{m\/s}^2$: $0^2 = u^2 + 2(-9.8)(20) \\implies 0 = u^2 - 392 \\implies u^2 = 392 \\implies u \\approx 19.8 \\, \\text{m\/s}$. Both approximations are closest to 20 m\/s among the given options.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-tennis-ball-is-thrown-in-the-vertically-upward-direction-and-the-bal\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-tennis-ball-is-thrown-in-the-vertically-upward-direction-and-the-bal\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-tennis-ball-is-thrown-in-the-vertically-upward-direction-and-the-bal\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-2\/"},{"@type":"ListItem","position":3,"name":"A tennis ball is thrown in the vertically upward direction and the bal"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88817","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=88817"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88817\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88817"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88817"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88817"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}