{"id":84,"date":"2018-01-14T10:21:44","date_gmt":"2018-01-14T02:21:44","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=84"},"modified":"2018-01-14T10:21:44","modified_gmt":"2018-01-14T02:21:44","slug":"solving-projectile-motion-problems-determine-final-velocity","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/solving-projectile-motion-problems-determine-final-velocity.htm","title":{"rendered":"Determine final velocity of projectile motion","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. <\/span><\/span>A kicked football leaves the ground at an angle \u03b8 = 30<\/span><\/span>o<\/span><\/span><\/sup> to the horizontal with an initial speed of 14 m\/s. Calculate the final velocity before the ball hits the ground.<\/span><\/span><\/p>\n

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

Angle (<\/span><\/span>\u03b8<\/span><\/span>) = 3<\/span><\/span>0<\/span><\/span>o<\/span><\/span><\/sup><\/p>\n

Initial velocity<\/a> (v<\/span><\/span>o<\/span><\/sub>) = 14 m\/s<\/span><\/span><\/p>\n

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

Wanted :<\/u><\/span><\/span> Final velocity before the ball hits the ground<\/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 = (14 m\/s)(cos 30<\/span><\/span>o<\/span><\/span><\/sup>) = (14 m\/s)(0.5<\/span><\/span>\u221a<\/span><\/span>3<\/span><\/span>) = 7<\/span><\/span>\u221a<\/span><\/span>3 <\/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 = (14 m\/s)(sin 30<\/span><\/span>o<\/span><\/span><\/sup>) = (14 m\/s)(0.5) = 7 m\/s<\/span><\/span><\/p>\n

Final velocity 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>) = 7 m\/s (positive upward)<\/span><\/span><\/p>\n

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

Height (h) = 0 (object back to initial position)<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Final velocity (v<\/span><\/span>t<\/span><\/sub>)<\/span><\/span><\/p>\n

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

v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> = v<\/span><\/span>o<\/span><\/span><\/sub>2 <\/span><\/span><\/sup>+ 2 g h = 7<\/span><\/span>2 <\/span><\/span><\/sup>+ 2(-10)(0) = 49 \u2013 0 = 49 <\/span><\/span><\/p>\n

v<\/span><\/span>t <\/span><\/sub>= <\/span><\/span>\u221a49 = 7 m\/s<\/span><\/span><\/p>\n

Final velocity at horizontal direction<\/b><\/span><\/span><\/p>\n

Initial velocity at horizontal direction is <\/span><\/span>7<\/span><\/span>\u221a<\/span><\/span>3 <\/span><\/span>m\/s. Velocity is constant so that final velocity is same as initial velocity. <\/span><\/span><\/p>\n

Final velocity before the object hits the ground<\/b><\/span><\/span><\/p>\n

\"Solving<\/p>\n

2. <\/span><\/span>A body is projected upward at an angle of 30<\/span><\/span>o <\/span><\/span><\/sup>with the horizontal from a building 5 meter high. Its initial speed is 10 m\/s. Calculate final velocity before the object hits the ground! 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>) = 30<\/span><\/span>o<\/span><\/span><\/sup><\/p>\n

Initial height (h<\/span><\/span>o<\/span><\/sub>) = 5 meters<\/span><\/span><\/p>\n

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

Solution :<\/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 = (10 m\/s)(cos 30<\/span><\/span>o<\/span><\/span><\/sup>) = (10 m\/s)(0.5<\/span><\/span>\u221a<\/span><\/span>3<\/span><\/span>) = 5<\/span><\/span>\u221a<\/span><\/span>3 <\/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 = (10 m\/s)(sin 30<\/span><\/span>o<\/span><\/span><\/sup>) = (10 m\/s)(0.5) = 5 m\/s<\/span><\/span><\/p>\n

Final velocity at vertical direction<\/b><\/span><\/span><\/p>\n

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

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = 5 m\/s (positive upward)<\/span><\/span><\/p>\n

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

Height (h) = -5 m (<\/span><\/span>negative because the ground is below the initial height)<\/span><\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Final velocity (v<\/span><\/span>t<\/span><\/span><\/sub>)<\/span><\/span><\/p>\n

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

v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> = v<\/span><\/span>o<\/span><\/span><\/sub>2 <\/span><\/span><\/sup>+ 2 g h = 5<\/span><\/span>2 <\/span><\/span><\/sup>+ 2(-10)(-5) = 25 + 100 = 125<\/span><\/span><\/p>\n

v<\/span><\/span>t<\/sub> <\/span><\/span>= <\/span><\/span>\u221a125 m\/s<\/span><\/span><\/p>\n

Final velocity at horizontal direction<\/b><\/span><\/span><\/p>\n

Final velocity at horizontal direction is <\/span><\/span>5<\/span><\/span>\u221a3 <\/span><\/span>m\/s. <\/span><\/span><\/p>\n

Final velocity <\/b><\/span><\/span><\/p>\n

\"Solving<\/p>\n

3. <\/span><\/span>A small ball projected horizontally with initial velocity v<\/span><\/span>o<\/span><\/sub> = 8 m\/s from a building 12 meters high. Calculate final velocity before ball hits ground<\/span><\/span>! Acceleration of gravity is 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

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

Height (h) = 12 meters<\/span><\/span><\/p>\n

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = 8 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> Final velocity (vt<\/sub>)<\/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> = 8 m\/s <\/span><\/span><\/p>\n

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

v<\/span><\/span>oy <\/span><\/sub>= 0 m\/s<\/span><\/span><\/p>\n

Final velocity at vertical direction<\/b><\/span><\/span><\/p>\n

calculated using equation of free fall motion<\/a>.
\n<\/span><\/span><\/p>\n

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

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

Height (h) = 12 m<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Final velocity (v<\/span><\/span>t<\/span><\/sub>)<\/span><\/span><\/p>\n

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

v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> = 2 g h = 2(10)(12) = 240<\/span><\/span><\/p>\n

v<\/span><\/span>t <\/span><\/sub>= <\/span><\/span>\u221a240 m\/s<\/span><\/span><\/p>\n

Final velocity at horizontal direction<\/b><\/span><\/span><\/p>\n

Initial velocity at the horizontal direction is 8 m\/s. Velocity is constant so that the initial velocity equals the final velocity. So final velocity at horizontal direction is 8 m\/s.<\/span><\/span><\/span><\/p>\n

Final velocity <\/b><\/span><\/span><\/p>\n

\"Solving<\/p>\n

[wpdm_package id=’534′]<\/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":"

    1. A kicked football leaves the ground at an angle \u03b8 = 30o to the horizontal with an initial speed of 14 m\/s. Calculate the final velocity before the ball hits the ground. Known : Angle (\u03b8) = 30o Initial velocity (vo) = 14 m\/s Acceleration of gravity (g) = 10 m\/s2 Wanted : Final … 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 final velocity 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-84","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\/84","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=84"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/84\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=84"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=84"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=84"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}