{"id":265426,"date":"2025-07-22T06:04:07","date_gmt":"2025-07-22T06:04:07","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=265426"},"modified":"2025-07-22T06:04:12","modified_gmt":"2025-07-22T06:04:12","slug":"an-object-is-shot-vertically-from-top-of-a-building-60-m-above-the-ground-with-an-initial-speed-vo27-m-s-at-point-o","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/22\/an-object-is-shot-vertically-from-top-of-a-building-60-m-above-the-ground-with-an-initial-speed-vo27-m-s-at-point-o\/","title":{"rendered":"An object is shot vertically from top of a building 60 m above the ground with an initial speed vo=27 m\/s (at point O)"},"content":{"rendered":"\n
An object is shot vertically from top of a building 60 m above the ground with an initial speed vo=27 m\/s (at point O). Define the ground as y=0 and call “UP” to be the positive y-axis. Assume no air. Responses should be based on Graphical analysis. No credit given for analytical solution a. Without using the kinematic equations determine the velocity at t=0 to t=4s. Show work and explain t (5) 0 1 2 3 4 V(m\/s) b. Based on the values in part a draw the velocity vs time graph for this motion velocity (m\/s) 30 20 10 0 -10 234 5 -20 -30 time (s) c. Based on the table or graph determine how long does it take to reach the maximum height? Show work and explain.<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Let’s break down the problem step by step and focus on graphical analysis.<\/p>\n\n\n\n
a. Determine the velocity at different times (t = 0, 1, 2, 3, 4 seconds):<\/strong><\/h3>\n\n\n\n
Since the object is shot vertically upward from the building at v0=27\u2009m\/sv_0 = 27 \\, \\text{m\/s}v0\u200b=27m\/s, and the motion is under gravity, we know the object will slow down until it stops momentarily (at the maximum height) and then start falling back. However, since we\u2019re using a graphical analysis, we can infer a lot of information just by understanding the motion and gravity\u2019s effects on velocity.<\/p>\n\n\n\n
\n
Initial velocity (at t=0t = 0t=0):<\/strong> Since the object is shot upwards with a velocity of v0=27\u2009m\/sv_0 = 27 \\, \\text{m\/s}v0\u200b=27m\/s, at t=0t = 0t=0, the velocity is 27\u2009m\/s27 \\, \\text{m\/s}27m\/s. So the velocity at t=0t = 0t=0 is 27 m\/s.<\/li>\n\n\n\n
At t=1st = 1st=1s:<\/strong> The object will experience a constant downward acceleration due to gravity, which is approximately 9.8\u2009m\/s29.8 \\, \\text{m\/s}^29.8m\/s2. So after 1 second, the velocity will decrease by 9.8\u2009m\/s9.8 \\, \\text{m\/s}9.8m\/s. Therefore, at t=1\u2009st = 1 \\, \\text{s}t=1s, the velocity is approximately v1=27\u22129.8=17.2\u2009m\/sv_1 = 27 – 9.8 = 17.2 \\, \\text{m\/s}v1\u200b=27\u22129.8=17.2m\/s.<\/li>\n\n\n\n
At t=2st = 2st=2s:<\/strong> After another second, the velocity will decrease by another 9.8\u2009m\/s9.8 \\, \\text{m\/s}9.8m\/s, so at t=2\u2009st = 2 \\, \\text{s}t=2s, the velocity is v2=17.2\u22129.8=7.4\u2009m\/sv_2 = 17.2 – 9.8 = 7.4 \\, \\text{m\/s}v2\u200b=17.2\u22129.8=7.4m\/s.<\/li>\n\n\n\n
At t=3st = 3st=3s:<\/strong> After 3 seconds, the velocity will decrease further by 9.8\u2009m\/s9.8 \\, \\text{m\/s}9.8m\/s, so at t=3\u2009st = 3 \\, \\text{s}t=3s, the velocity is v3=7.4\u22129.8=\u22122.4\u2009m\/sv_3 = 7.4 – 9.8 = -2.4 \\, \\text{m\/s}v3\u200b=7.4\u22129.8=\u22122.4m\/s. This negative velocity indicates that the object is now moving downwards after reaching its highest point.<\/li>\n\n\n\n
At t=4st = 4st=4s:<\/strong> Finally, after 4 seconds, the object will have accelerated downward further, so the velocity is v4=\u22122.4\u22129.8=\u221212.2\u2009m\/sv_4 = -2.4 – 9.8 = -12.2 \\, \\text{m\/s}v4\u200b=\u22122.4\u22129.8=\u221212.2m\/s.<\/li>\n<\/ul>\n\n\n\n
b. Draw the Velocity vs Time graph:<\/strong><\/h3>\n\n\n\n
We can plot the velocity values from the table onto a graph. The y-axis represents velocity (in m\/s) and the x-axis represents time (in seconds). Here’s how the graph would look:<\/p>\n\n\n\n
\n
At t=0t = 0t=0,<\/strong> the velocity is +27\u2009m\/s+27 \\, \\text{m\/s}+27m\/s, which starts the curve at the top of the graph.<\/li>\n\n\n\n
At t=1st = 1st=1s,<\/strong> the velocity drops to +17.2\u2009m\/s+17.2 \\, \\text{m\/s}+17.2m\/s.<\/li>\n\n\n\n
At t=2st = 2st=2s,<\/strong> the velocity further drops to +7.4\u2009m\/s+7.4 \\, \\text{m\/s}+7.4m\/s.<\/li>\n\n\n\n
At t=3st = 3st=3s,<\/strong> the velocity becomes negative at \u22122.4\u2009m\/s -2.4 \\, \\text{m\/s}\u22122.4m\/s, indicating downward motion.<\/li>\n\n\n\n
At t=4st = 4st=4s,<\/strong> the velocity continues in the downward direction to \u221212.2\u2009m\/s -12.2 \\, \\text{m\/s}\u221212.2m\/s.<\/li>\n<\/ul>\n\n\n\n
The graph will be a straight line with a negative slope (because gravity is acting uniformly) and crossing the time-axis at the point where the object reaches its highest point (maximum height).<\/p>\n\n\n\n
c. Determine how long it takes to reach the maximum height:<\/strong><\/h3>\n\n\n\n
The maximum height occurs when the velocity becomes zero. Based on the velocity-time graph, the object starts at a velocity of 27\u2009m\/s27 \\, \\text{m\/s}27m\/s and slows down at a constant rate due to gravity. The velocity reaches zero when the object stops moving upwards and starts falling.<\/p>\n\n\n\n
From the table, we see that the velocity decreases linearly with time. The object reaches v=0\u2009m\/sv = 0 \\, \\text{m\/s}v=0m\/s between t=2\u2009st = 2 \\, \\text{s}t=2s and t=3\u2009st = 3 \\, \\text{s}t=3s. Since the object\u2019s velocity is decreasing at a constant rate, we can estimate the time to reach the maximum height by interpolation between the values for t=2st = 2st=2s and t=3st = 3st=3s.<\/p>\n\n\n\n
\n
At t=2\u2009st = 2 \\, \\text{s}t=2s, the velocity is 7.4\u2009m\/s7.4 \\, \\text{m\/s}7.4m\/s, and at t=3\u2009st = 3 \\, \\text{s}t=3s, the velocity is \u22122.4\u2009m\/s-2.4 \\, \\text{m\/s}\u22122.4m\/s.<\/li>\n\n\n\n
Since the velocity decreases linearly, it takes approximately 2.7\u2009seconds2.7 \\, \\text{seconds}2.7seconds to reach the maximum height, where the velocity is zero.<\/li>\n<\/ul>\n\n\n\n
Thus, the time to reach the maximum height is approximately 2.7 seconds<\/strong>.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
An object is shot vertically from top of a building 60 m above the ground with an initial speed vo=27 m\/s (at point O). Define the ground as y=0 and call “UP” to be the positive y-axis. Assume no air. Responses should be based on Graphical analysis. No credit given for analytical solution a. Without […]<\/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-265426","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/265426","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=265426"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/265426\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=265426"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=265426"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=265426"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-1455.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"