{"id":76,"date":"2018-01-14T05:37:07","date_gmt":"2018-01-13T21:37:07","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=76"},"modified":"2018-01-14T05:37:07","modified_gmt":"2018-01-13T21:37:07","slug":"solving-projectile-motion-problems-determine-the-position-of-an-object","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/solving-projectile-motion-problems-determine-the-position-of-an-object.htm","title":{"rendered":"Determine the position of an object in projectile motion","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Solved problems in projectile motion<\/a> – determine the position of an object<\/a><\/p>\n

1. A body is projected upward at an angle of 60<\/span><\/span>o <\/span><\/span><\/sup>to <\/span><\/span>the horizontal with an initial speed of 12 m\/s. Determine the position of the object after moving for 1 second! Acceleration of gravity<\/a> 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><\/a><\/p>\n

Initial velocity<\/a> (v<\/span><\/span>o<\/span><\/sub>) = 12 m\/s<\/span><\/span><\/p>\n

Time interval (t) = 1 second<\/span><\/span><\/p>\n

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

Wanted :<\/u><\/span><\/span> Object position after moving during 1 second<\/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 = (12 m\/s)(cos 60<\/span><\/span>o<\/span><\/span><\/sup>) = (12 m\/s)(0.5) = 6 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 = (12 m\/s)(sin 60<\/span><\/span>o<\/span><\/span><\/sup>) = (12 m\/s)(0.5<\/span><\/span>\u221a<\/span><\/span>3<\/span><\/span>) = 6<\/span><\/span>\u221a<\/span><\/span>3 <\/span><\/span>m\/s<\/span><\/span><\/p>\n

Object position at horizontal direction:<\/b><\/span><\/span><\/p>\n

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

Horizontal component of velocity (v<\/span><\/span>x<\/span><\/sub>) = 6 m\/s<\/span><\/span><\/p>\n

Time interval (t) = 1 second<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> horizontal range (x)<\/span><\/span><\/p>\n

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

6 meters \/ second means the ball moves as far as 6 meters every 1 second. The distance of the ball after moving for 1 second is 6 meters. So the position of the ball in the horizontal direction is 6 meters.<\/span><\/span><\/span><\/p>\n

Object position at vertical direction :<\/b><\/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>6<\/span><\/span>\u221a<\/span><\/span>3 <\/span><\/span>m\/s (positive upward) <\/span><\/span><\/p>\n

Time interval (t) = 1 second<\/span><\/span><\/p>\n

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

Wanted :<\/u><\/span><\/span> height after moving during 1 second <\/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> = (<\/span><\/span>6<\/span><\/span>\u221a<\/span><\/span>3)(1) <\/span><\/span>+ 1\/2 (-10)(1<\/span><\/span>2<\/span><\/span><\/sup>) = <\/span><\/span>6<\/span><\/span>\u221a<\/span><\/span>3 + (-5)(1) = <\/span><\/span>6<\/span><\/span>\u221a<\/span><\/span>3 \u2013 5 = 6(1.7) \u2013 5 = 10.2 \u2013 5 = 5.2 meters. <\/span><\/span><\/p>\n

Object position after moving for 1 second :<\/b><\/span><\/span><\/p>\n

Horizontal displacement (x) = 6 meters<\/span><\/span><\/p>\n

Vertical displacement (y) = 5.2 meters<\/span><\/span><\/p>\n

<\/p>\n

2. A body is projected upward at an angle of 30<\/span><\/span>o <\/span><\/span><\/sup>to <\/span><\/span>the horizontal from a building 20 meters high. Its initial speed is 50 m\/s. Calculate the vertical displacement after the body moving for 1 second! 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>30<\/span><\/span>o<\/span><\/span><\/sup><\/p>\n

Initial height (h<\/span><\/span>o<\/span><\/sub>) = 20 meters<\/span><\/span><\/p>\n

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = <\/span><\/span>50 m\/s<\/span><\/span><\/p>\n

Time interval (t) = 1 second<\/span><\/span><\/p>\n

Acceleration of gravity (g) = 1<\/span><\/span>0 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Wanted :<\/u><\/span><\/span> Height (h)<\/span><\/span><\/p>\n

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

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

v<\/span><\/span>oy<\/sub> <\/span><\/span>= v<\/span><\/span>o<\/span><\/sub> sin \u03b8 = (50 m\/s)(sin 30<\/span><\/span>o<\/span><\/span><\/sup>) = (50 m\/s)(0.5) = 25<\/span><\/span> m\/s<\/span><\/span><\/p>\n

Height :<\/b><\/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>25<\/span><\/span> m\/s (positive upward)<\/span><\/span><\/p>\n

Time interval (t) = 1 second<\/span><\/span><\/p>\n

Acceleration of gravity (g) = -1<\/span><\/span>0 m\/s<\/span><\/span>2<\/span><\/span><\/sup> (negative downward)<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Height (h)<\/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>= (25)(1) + 1\/2 (-10)(1<\/span><\/span>2<\/span><\/span><\/sup>) = 25 + (-5)(1) = 25 \u2013 5 = 20 meters.<\/span><\/span><\/p>\n

The height of the body after moving for 1 second is 20 meters above where the body is <\/span><\/span><\/span>projected <\/span><\/span><\/span>or 40 meters above ground.<\/span><\/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 meters high. Calculate the ball displacement after moving 1 second<\/span><\/span>! Acceleration of gravity is 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

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

Initial height (h) = 10 meters<\/span><\/span><\/p>\n

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

Time interval (t) = 1 second<\/span><\/span><\/p>\n

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

Wanted:<\/u><\/span><\/span> Position of the ball after moving 1 second!<\/span><\/span><\/p>\n

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

\"SolvingHorizontal displacement :<\/b><\/span><\/span><\/p>\n

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

Horizontal component of velocity (v<\/span><\/span>x<\/span><\/sub>) = 10 m\/s<\/span><\/span><\/p>\n

Time interval (t) = 1 second<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> Position of the object<\/span><\/span><\/p>\n

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

10 meters\/second means the object move as far as 10 meters every 1 second. Displacement<\/a> after moving for 1 second is 10 meters. So horizontal displacement is 10 meters.<\/span><\/span><\/span><\/p>\n

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

Calculated as the free fall motion<\/a>.<\/i><\/span><\/span><\/p>\n

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

Time interval (t) = 1 second<\/span><\/span><\/p>\n

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

Wanted :<\/u><\/span><\/span> Height after moving during 1 second (h)<\/span><\/span><\/p>\n

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

h = 1\/2 g t<\/span><\/span>2<\/span><\/span><\/sup> = 1\/2 (10)(1<\/span><\/span>2<\/span><\/span><\/sup>) = <\/span><\/span>(5)(1) = 5 meters. <\/span><\/span><\/p>\n

After 1 second, the object falls as far as 5 meters. Height above the ground level = 10 meters – 5 meters = 5 meters.<\/span><\/span><\/span><\/p>\n

The position of the object after moving 1 second :<\/b><\/span><\/span><\/p>\n

Position of the object at horizontal direction<\/a> (x) = 10 meters<\/span><\/span><\/p>\n

The position of the object at vertical direction (y) = 5 meters<\/span><\/span><\/p>\n

<\/p>\n

[wpdm_package id=’532′]<\/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 the 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 position of an object 1. A body is projected upward at an angle of 60o to the horizontal with an initial speed of 12 m\/s. Determine the position of the object after moving for 1 second! Acceleration of gravity is 10 m\/s2. Known : Angle (\u03b8) = … 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 the position of an object in 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-76","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\/76","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=76"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/76\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=76"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=76"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=76"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}