{"id":90932,"date":"2025-06-01T10:41:28","date_gmt":"2025-06-01T10:41:28","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=90932"},"modified":"2025-06-01T10:41:28","modified_gmt":"2025-06-01T10:41:28","slug":"a-canon-shoots-a-ball-upwards-with-an-initial-speed-of-100-m-s-the-to","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-canon-shoots-a-ball-upwards-with-an-initial-speed-of-100-m-s-the-to\/","title":{"rendered":"A canon shoots a ball upwards with an initial speed of 100 m\/s. The to"},"content":{"rendered":"

A canon shoots a ball upwards with an initial speed of 100 m\/s. The total time of flight of the ball is 20 s before it hits the ground. The ball looses 70% of its speed after hitting the ground. Which among the following is the correct height that the ball will bounce up after its first bounce? (g=10 m\/s\u00b2)<\/p>\n

[amp_mcq option1=”100 m” option2=”70 m” option3=”50 m” option4=”45 m” correct=”option4″]<\/p>\n

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This question was previously asked in<\/div>\n
UPSC CAPF – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe ball is shot upwards with an initial speed of 100 m\/s. In vertical motion under gravity (g=10 m\/s\u00b2), the time taken to reach the maximum height is given by t_up = u\/g = 100\/10 = 10 s. The time taken to fall back to the ground from the maximum height is also 10 s. The total time of flight before the first bounce is t_up + t_down = 10 + 10 = 20 s, which matches the given information. The speed of the ball just before hitting the ground after falling from its peak height is equal to its initial projection speed, which is 100 m\/s (downwards). After hitting the ground, the ball loses 70% of its speed. So, the speed after the first bounce is (100% – 70%) of 100 m\/s = 30% of 100 m\/s = 0.30 * 100 m\/s = 30 m\/s (upwards). To find the height the ball bounces up, we use the equation v\u00b2 = u\u00b2 + 2as, where the final velocity at the peak height is v=0, the initial velocity after bounce is u=30 m\/s, and the acceleration is a=-g=-10 m\/s\u00b2. So, 0\u00b2 = 30\u00b2 + 2 * (-10) * h. This simplifies to 0 = 900 – 20h, so 20h = 900, which gives h = 900\/20 = 45 meters.
\n<\/section>\n
\nUnderstanding projectile motion under gravity, calculating time of flight, speed before impact, calculating speed after an inelastic collision (loss of speed), and calculating the maximum height reached with a new initial velocity.
\n<\/section>\n
\nThe collision with the ground is inelastic as the ball loses speed. The coefficient of restitution (e) for this bounce would be the ratio of the speed after bounce to the speed before bounce, i.e., e = 30\/100 = 0.3. The maximum height reached after a bounce with speed v is given by h = v\u00b2 \/ (2g).
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A canon shoots a ball upwards with an initial speed of 100 m\/s. The total time of flight of the ball is 20 s before it hits the ground. The ball looses 70% of its speed after hitting the ground. Which among the following is the correct height that the ball will bounce up after … <\/p>\n

Detailed SolutionA canon shoots a ball upwards with an initial speed of 100 m\/s. The to<\/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":[1085],"tags":[1103,1421,1128],"class_list":["post-90932","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1103","tag-motion-under-gravity","tag-physics","no-featured-image-padding"],"yoast_head":"\nA canon shoots a ball upwards with an initial speed of 100 m\/s. The to<\/title>\n<meta name=\"description\" content=\"The ball is shot upwards with an initial speed of 100 m\/s. In vertical motion under gravity (g=10 m\/s\u00b2), the time taken to reach the maximum height is given by t_up = u\/g = 100\/10 = 10 s. The time taken to fall back to the ground from the maximum height is also 10 s. The total time of flight before the first bounce is t_up + t_down = 10 + 10 = 20 s, which matches the given information. The speed of the ball just before hitting the ground after falling from its peak height is equal to its initial projection speed, which is 100 m\/s (downwards). After hitting the ground, the ball loses 70% of its speed. So, the speed after the first bounce is (100% - 70%) of 100 m\/s = 30% of 100 m\/s = 0.30 * 100 m\/s = 30 m\/s (upwards). To find the height the ball bounces up, we use the equation v\u00b2 = u\u00b2 + 2as, where the final velocity at the peak height is v=0, the initial velocity after bounce is u=30 m\/s, and the acceleration is a=-g=-10 m\/s\u00b2. So, 0\u00b2 = 30\u00b2 + 2 * (-10) * h. This simplifies to 0 = 900 - 20h, so 20h = 900, which gives h = 900\/20 = 45 meters. Understanding projectile motion under gravity, calculating time of flight, speed before impact, calculating speed after an inelastic collision (loss of speed), and calculating the maximum height reached with a new initial velocity.\" \/>\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-canon-shoots-a-ball-upwards-with-an-initial-speed-of-100-m-s-the-to\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A canon shoots a ball upwards with an initial speed of 100 m\/s. The to\" \/>\n<meta property=\"og:description\" content=\"The ball is shot upwards with an initial speed of 100 m\/s. In vertical motion under gravity (g=10 m\/s\u00b2), the time taken to reach the maximum height is given by t_up = u\/g = 100\/10 = 10 s. The time taken to fall back to the ground from the maximum height is also 10 s. The total time of flight before the first bounce is t_up + t_down = 10 + 10 = 20 s, which matches the given information. The speed of the ball just before hitting the ground after falling from its peak height is equal to its initial projection speed, which is 100 m\/s (downwards). After hitting the ground, the ball loses 70% of its speed. So, the speed after the first bounce is (100% - 70%) of 100 m\/s = 30% of 100 m\/s = 0.30 * 100 m\/s = 30 m\/s (upwards). To find the height the ball bounces up, we use the equation v\u00b2 = u\u00b2 + 2as, where the final velocity at the peak height is v=0, the initial velocity after bounce is u=30 m\/s, and the acceleration is a=-g=-10 m\/s\u00b2. So, 0\u00b2 = 30\u00b2 + 2 * (-10) * h. This simplifies to 0 = 900 - 20h, so 20h = 900, which gives h = 900\/20 = 45 meters. Understanding projectile motion under gravity, calculating time of flight, speed before impact, calculating speed after an inelastic collision (loss of speed), and calculating the maximum height reached with a new initial velocity.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-canon-shoots-a-ball-upwards-with-an-initial-speed-of-100-m-s-the-to\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T10:41:28+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=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"A canon shoots a ball upwards with an initial speed of 100 m\/s. The to","description":"The ball is shot upwards with an initial speed of 100 m\/s. In vertical motion under gravity (g=10 m\/s\u00b2), the time taken to reach the maximum height is given by t_up = u\/g = 100\/10 = 10 s. The time taken to fall back to the ground from the maximum height is also 10 s. The total time of flight before the first bounce is t_up + t_down = 10 + 10 = 20 s, which matches the given information. The speed of the ball just before hitting the ground after falling from its peak height is equal to its initial projection speed, which is 100 m\/s (downwards). After hitting the ground, the ball loses 70% of its speed. So, the speed after the first bounce is (100% - 70%) of 100 m\/s = 30% of 100 m\/s = 0.30 * 100 m\/s = 30 m\/s (upwards). To find the height the ball bounces up, we use the equation v\u00b2 = u\u00b2 + 2as, where the final velocity at the peak height is v=0, the initial velocity after bounce is u=30 m\/s, and the acceleration is a=-g=-10 m\/s\u00b2. So, 0\u00b2 = 30\u00b2 + 2 * (-10) * h. This simplifies to 0 = 900 - 20h, so 20h = 900, which gives h = 900\/20 = 45 meters. Understanding projectile motion under gravity, calculating time of flight, speed before impact, calculating speed after an inelastic collision (loss of speed), and calculating the maximum height reached with a new initial velocity.","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-canon-shoots-a-ball-upwards-with-an-initial-speed-of-100-m-s-the-to\/","og_locale":"en_US","og_type":"article","og_title":"A canon shoots a ball upwards with an initial speed of 100 m\/s. The to","og_description":"The ball is shot upwards with an initial speed of 100 m\/s. In vertical motion under gravity (g=10 m\/s\u00b2), the time taken to reach the maximum height is given by t_up = u\/g = 100\/10 = 10 s. The time taken to fall back to the ground from the maximum height is also 10 s. The total time of flight before the first bounce is t_up + t_down = 10 + 10 = 20 s, which matches the given information. The speed of the ball just before hitting the ground after falling from its peak height is equal to its initial projection speed, which is 100 m\/s (downwards). After hitting the ground, the ball loses 70% of its speed. So, the speed after the first bounce is (100% - 70%) of 100 m\/s = 30% of 100 m\/s = 0.30 * 100 m\/s = 30 m\/s (upwards). To find the height the ball bounces up, we use the equation v\u00b2 = u\u00b2 + 2as, where the final velocity at the peak height is v=0, the initial velocity after bounce is u=30 m\/s, and the acceleration is a=-g=-10 m\/s\u00b2. So, 0\u00b2 = 30\u00b2 + 2 * (-10) * h. This simplifies to 0 = 900 - 20h, so 20h = 900, which gives h = 900\/20 = 45 meters. Understanding projectile motion under gravity, calculating time of flight, speed before impact, calculating speed after an inelastic collision (loss of speed), and calculating the maximum height reached with a new initial velocity.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-canon-shoots-a-ball-upwards-with-an-initial-speed-of-100-m-s-the-to\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T10:41:28+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-canon-shoots-a-ball-upwards-with-an-initial-speed-of-100-m-s-the-to\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-canon-shoots-a-ball-upwards-with-an-initial-speed-of-100-m-s-the-to\/","name":"A canon shoots a ball upwards with an initial speed of 100 m\/s. The to","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T10:41:28+00:00","dateModified":"2025-06-01T10:41:28+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The ball is shot upwards with an initial speed of 100 m\/s. In vertical motion under gravity (g=10 m\/s\u00b2), the time taken to reach the maximum height is given by t_up = u\/g = 100\/10 = 10 s. The time taken to fall back to the ground from the maximum height is also 10 s. The total time of flight before the first bounce is t_up + t_down = 10 + 10 = 20 s, which matches the given information. The speed of the ball just before hitting the ground after falling from its peak height is equal to its initial projection speed, which is 100 m\/s (downwards). After hitting the ground, the ball loses 70% of its speed. So, the speed after the first bounce is (100% - 70%) of 100 m\/s = 30% of 100 m\/s = 0.30 * 100 m\/s = 30 m\/s (upwards). To find the height the ball bounces up, we use the equation v\u00b2 = u\u00b2 + 2as, where the final velocity at the peak height is v=0, the initial velocity after bounce is u=30 m\/s, and the acceleration is a=-g=-10 m\/s\u00b2. So, 0\u00b2 = 30\u00b2 + 2 * (-10) * h. This simplifies to 0 = 900 - 20h, so 20h = 900, which gives h = 900\/20 = 45 meters. Understanding projectile motion under gravity, calculating time of flight, speed before impact, calculating speed after an inelastic collision (loss of speed), and calculating the maximum height reached with a new initial velocity.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-canon-shoots-a-ball-upwards-with-an-initial-speed-of-100-m-s-the-to\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-canon-shoots-a-ball-upwards-with-an-initial-speed-of-100-m-s-the-to\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-canon-shoots-a-ball-upwards-with-an-initial-speed-of-100-m-s-the-to\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CAPF","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-capf\/"},{"@type":"ListItem","position":3,"name":"A canon shoots a ball upwards with an initial speed of 100 m\/s. 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