{"id":58,"date":"2018-01-13T16:10:24","date_gmt":"2018-01-13T08:10:24","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=58"},"modified":"2018-01-13T16:10:24","modified_gmt":"2018-01-13T08:10:24","slug":"solving-projectile-motion-problems-determine-horizontal-displacement","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/solving-projectile-motion-problems-determine-horizontal-displacement.htm","title":{"rendered":"Determine horizontal displacement of projectile motion","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Solved problems in projectile motion<\/a> – determine the horizontal displacement<\/a><\/p>\n

1. <\/span><\/span>A kicked football leaves the ground at an angle \u03b8 = 60<\/span><\/span>o <\/span><\/span><\/sup>with the horizontal has an initial speed of 16 m\/s. How long will it be before the ball hits the ground?<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Angle (<\/span><\/span>\u03b8<\/span><\/span>) = <\/span><\/span>60<\/span><\/span>o<\/span><\/span><\/sup><\/p>\n

Initial speed (v<\/span><\/span>o<\/span><\/sub>) = <\/span><\/span>16 m\/s<\/span><\/span><\/p>\n

Acceleration of gravity (g) = 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Wanted :<\/u><\/span><\/span> Horizontal displacement (x)<\/span><\/span><\/p>\n

\"SolvingSolution :<\/u><\/span><\/span><\/p>\n

Horizontal component of initial velocity :<\/span><\/span><\/p>\n

v<\/span><\/span>ox <\/span><\/sub>= v<\/span><\/span>o<\/span><\/sub> cos \u03b8 = (16 m\/s)(cos 60<\/span><\/span>o<\/span><\/span><\/sup>) = (16 m\/s)(0.5) = 8<\/span><\/span> m\/s<\/span><\/span><\/p>\n

Vertical component of initial velocity :<\/span><\/span><\/p>\n

v<\/span><\/span>oy <\/span><\/sub>= v<\/span><\/span>o<\/span><\/sub> sin \u03b8 = (16 m\/s)(sin 60<\/span><\/span>o<\/span><\/span><\/sup>) = (16 m\/s)(0.5<\/span><\/span>\u221a<\/span><\/span>3<\/span><\/span>) = 8<\/span><\/span>\u221a<\/span><\/span>3 <\/span><\/span>m\/s<\/span><\/span><\/p>\n

Projectile motion<\/a> could be understood by analyzing the horizontal and vertical component of the motion separately. The x motion occurs at constant velocity and the y motion occurs at constant acceleration of gravity. <\/span><\/span><\/p>\n

Time in the air<\/b><\/span><\/span><\/p>\n

The time it stays in the air is determined by the y motion. We first find the time using the y motion and then use this time value in the x equations (constant velocity<\/a> equation).<\/span><\/span><\/p>\n

Choose upward direction as positive and downward direction as negative.<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = <\/span><\/span>8<\/span><\/span>\u221a<\/span><\/span>3 <\/span><\/span>m\/s<\/span><\/span> (v<\/span><\/span>o<\/span><\/sub> upward)<\/span><\/span><\/p>\n

Acceleration of gravity (g) = -10 m\/s<\/span><\/span>2 <\/span><\/span><\/sup>(g downward)<\/span><\/span><\/p>\n

Height (h) = 0 (ball is back to the same position)<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Time in air<\/span><\/span><\/p>\n

Solution :
\n<\/u><\/span><\/span><\/p>\n

h = v<\/span><\/span>o<\/span><\/sub> t + 1\/2 g t<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

0 = (<\/span><\/span>8<\/span><\/span>\u221a<\/span><\/span>3)<\/span><\/span> t + 1\/2 (-10) t<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

0 = <\/span><\/span>8<\/span><\/span>\u221a<\/span><\/span>3<\/span><\/span> t – 5 t<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

8<\/span><\/span>\u221a<\/span><\/span>3<\/span><\/span> t = 5 t<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

8 <\/span><\/span>(1.7) <\/span><\/span>= 5 t<\/span><\/span><\/p>\n

14 <\/span><\/span>= 5 t<\/span><\/span><\/p>\n

t = <\/span><\/span>14 <\/span><\/span>\/ 5 = 2.8 seconds<\/span><\/span><\/p>\n

Horizontal displacement<\/b><\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Velocity<\/a> (v) = 8 m\/s<\/span><\/span><\/p>\n

Time interval (t) = 2.8 seconds<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Displacement<\/span><\/span><\/p>\n

Solution :
\n<\/u><\/span><\/span><\/p>\n

x = v t = (8 m\/s)(2.8 s) = 22.4 meters<\/span><\/span><\/p>\n

Horizontal displacement is 22.4 meters. <\/span><\/span><\/p>\n

<\/p>\n

2. A body is projected upward at angle of 60<\/span><\/span>o <\/span><\/span><\/sup>with the horizontal from a building 50 meter high. It’s initial speed is 30 m\/s. Calculate the horizontal displacement! Acceleration of gravity is 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup>.<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Angle (<\/span><\/span>\u03b8<\/span><\/span>) = <\/span><\/span>60<\/span><\/span>o<\/span><\/span><\/sup><\/p>\n

High (h) = 15 m<\/span><\/span><\/p>\n

Initial speed (v<\/span><\/span>o<\/span><\/sub>) = 30 <\/span><\/span>m\/s<\/span><\/span><\/p>\n

Acceleration of gravity (g) = 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Wanted :<\/u><\/span><\/span> x <\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

\"SolvingHorizontal component of initial velocity ::<\/span><\/span><\/p>\n

v<\/span><\/span>ox <\/span><\/sub>= v<\/span><\/span>o<\/span><\/sub> cos \u03b8 = (30 m\/s)(cos 60<\/span><\/span>o<\/span><\/span><\/sup>) = (30 m\/s)(0.5) = 15 m\/s<\/span><\/span><\/p>\n

Vertical component of initial velocity :<\/span><\/span><\/p>\n

v<\/span><\/span>oy <\/span><\/sub>= v<\/span><\/span>o<\/span><\/sub> sin \u03b8 = (30 m\/s)(sin 60<\/span><\/span>o<\/span><\/span><\/sup>) = (30 m\/s)(0.5<\/span><\/span>\u221a<\/span><\/span>3<\/span><\/span>) = 15<\/span><\/span>\u221a<\/span><\/span>3 <\/span><\/span>m\/s<\/span><\/span><\/p>\n

Time in the air<\/b><\/span><\/span><\/p>\n

We first find the time using the y motion and then use this time value in the x equations (constant velocity equation). Choose upward as positive and downward as negative.<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = <\/span><\/span>15<\/span><\/span>\u221a<\/span><\/span>3 <\/span><\/span>m\/s<\/span><\/span> (positive upward)<\/span><\/span><\/p>\n

Acceleration of gravity (g) = -10 m\/s<\/span><\/span>2 <\/span><\/span><\/sup>(negative downward)<\/span><\/span><\/p>\n

High (h) = -50 (Ground 50 meter below the initial position)<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> t<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

h = v<\/span><\/span>o<\/span><\/sub> t + 1\/2 g t<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

-50 = (<\/span><\/span>15<\/span><\/span>\u221a<\/span><\/span>3<\/span><\/span>)<\/span><\/span> t + 1\/2 (-10) t<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

-50 = <\/span><\/span>15<\/span><\/span>\u221a<\/span><\/span>3 <\/span><\/span>t – 5 t<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

5 t<\/span><\/span>2<\/span><\/span><\/sup> – <\/span><\/span>15<\/span><\/span>\u221a<\/span><\/span>3 <\/span><\/span>t \u2013 50 = 0<\/span><\/span><\/p>\n

Calculate time using this formula :<\/span><\/span><\/p>\n

a = 5, b = –<\/span><\/span>15<\/span><\/span>\u221a<\/span><\/span>3, c = –<\/span><\/span>50<\/span><\/span><\/p>\n

\"Solving<\/p>\n

Time in the air is 6.7 seconds.<\/span><\/span><\/p>\n

Horizontal displacement :<\/b><\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Velocity (v) = 15 m\/s<\/span><\/span><\/p>\n

Time interval (t) = 6.7 seconds<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> displacement<\/span><\/span><\/a><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

s = v t = (15 m\/s)(6.7 s) = 100.5 meters<\/span><\/span><\/p>\n

Horizontal displacement is 100.5 meters. <\/span><\/span><\/p>\n

<\/p>\n

3. <\/span><\/span>A small ball projected horizontally with initial velocity v<\/span><\/span>o<\/span><\/sub> = 10 m\/s from a building 10 meter high. Calculate the horizontal displacement<\/span><\/span>! Acceleration of gravity is 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

High (h) = 10 m<\/span><\/span><\/p>\n

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = 10 <\/span><\/span>m\/s<\/span><\/span><\/p>\n

Acceleration of gravity<\/a> (g) = 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Wanted :<\/u><\/span><\/span> x<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

\"SolvingHorizontal component of initial velocity = initial velocity = 10 m\/s.<\/span><\/span><\/p>\n

Time in the air<\/b><\/span><\/span><\/p>\n

Time in air calculated using free fall motion<\/a> equation.<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Acceleration of gravity (g) = 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

High (h) = 10 meter<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> t<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

h = 1\/2 g t<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

10 = 1\/2 (10) t<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

10 = 5 t<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

t<\/span><\/span>2 <\/span><\/span><\/sup>= 10 \/ 5 = 2<\/span><\/span><\/p>\n

t = <\/span><\/span>\u221a2 = 1.4 seconds<\/span><\/span><\/p>\n

Horizontal displacement<\/b><\/span><\/span><\/p>\n

Horizontal displacement calculated using equation of motion at constant velocity<\/a>.<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Velocity (v) = 10 m\/s<\/span><\/span><\/p>\n

Time interval (t) = 1.4 seconds<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> x<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

s = v t = (10 m\/s)(1.4 s) = 14 meters<\/span><\/span><\/p>\n

Horizontal displacement is 14 meters. <\/span><\/span><\/p>\n

<\/p>\n

[wpdm_package id=’526′]<\/p>\n

[wpdm_package id=’536′]<\/p>\n

    \n
  1. Resolve initial velocity into horizontal and vertical components<\/a><\/li>\n
  2. Determine the horizontal displacement<\/a><\/li>\n
  3. Determine the maximum height <\/a><\/li>\n
  4. Determine the time interval<\/a><\/li>\n
  5. Determine the position of object<\/a><\/li>\n
  6. Determine the final velocity<\/a><\/li>\n<\/ol>\n

    <\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

    Solved problems in projectile motion – determine the horizontal displacement 1. A kicked football leaves the ground at an angle \u03b8 = 60o with the horizontal has an initial speed of 16 m\/s. How long will it be before the ball hits the ground? Known : Angle (\u03b8) = 60o Initial speed (vo) = 16 … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Determine horizontal displacement of projectile motion","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-58","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/58","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=58"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/58\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=58"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=58"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=58"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}