{"id":65,"date":"2018-01-13T16:18:24","date_gmt":"2018-01-13T08:18:24","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=65"},"modified":"2018-01-13T16:18:24","modified_gmt":"2018-01-13T08:18:24","slug":"solving-projectile-motion-problems-determine-the-maximum-height","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/solving-projectile-motion-problems-determine-the-maximum-height.htm","title":{"rendered":"Determine the maximum height of projectile motion","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Solved problems in projectile motion<\/a> – determine the maximum height<\/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 10 m\/s. Calculate the maximum height! 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 (\u03b8) = 60<\/span><\/span>o<\/span><\/span><\/sup><\/p>\n

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

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

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

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

sin 60<\/span><\/span>o<\/span><\/span><\/sup> = v<\/span><\/span>oy<\/span><\/sub> \/ v<\/span><\/span>o<\/span><\/sub><\/p>\n

v<\/span><\/span>oy<\/span><\/sub> = v<\/span><\/span>o<\/span><\/sub> sin 60<\/span><\/span>o <\/span><\/span><\/sup>= (10)(sin 60<\/span><\/span>o<\/span><\/span><\/sup>) = (10)(0.5<\/span><\/span>\u221a<\/span><\/span>3<\/span><\/span>) = 5<\/span><\/span>\u221a<\/span><\/span>3 m\/s<\/span><\/span><\/p>\n

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

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

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

Vertical component of initial velocity (v<\/span><\/span>oy<\/span><\/sub>) = +<\/span><\/span>5<\/span><\/span>\u221a<\/span><\/span>3 m\/s <\/span><\/span>(positive upward)<\/i><\/span><\/span><\/p>\n

Final velocity at the maximum height (v<\/span><\/span>ty<\/span><\/sub>) = 0<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Maximum height (h)<\/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><\/sub>2<\/span><\/span><\/sup> + 2 g h<\/span><\/span><\/p>\n

0<\/span><\/span>2<\/span><\/span><\/sup> = <\/span><\/span>(<\/span><\/span>5<\/span><\/span>\u221a<\/span><\/span>3)<\/span><\/span>2<\/span><\/span><\/sup> + 2 (-10) h<\/span><\/span><\/p>\n

0 <\/span><\/span>= 2<\/span><\/span>5<\/span><\/span>(<\/span><\/span>3) <\/span><\/span>\u2013 20 h<\/span><\/span><\/p>\n

0 <\/span><\/span>= 75<\/span><\/span> \u2013 20 h<\/span><\/span><\/p>\n

75<\/span><\/span> =<\/span><\/span> 20 h<\/span><\/span><\/p>\n

h = 75 \/ 20<\/span><\/span><\/p>\n

h = 3.75 meter<\/span><\/span><\/p>\n

The maximum height is 3.75 meter.<\/span><\/span><\/p>\n

<\/p>\n

2. <\/span><\/span>A body is projected upward at angle of 30<\/span><\/span>o <\/span><\/span><\/sup>with the horizontal from a building 20 meter high. It’s initial speed is 4 m\/s. Calculate the maximum height! Acceleration of gravity is 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup>.<\/span><\/span><\/p>\n

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

Angle (\u03b8) = 30<\/span><\/span>o<\/span><\/span><\/sup><\/p>\n

Initial height (h) = 20 meter<\/span><\/span><\/p>\n

Initial velocity<\/a> (v<\/span><\/span>o<\/span><\/sub>) = 4 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> The maximum height (h)<\/span><\/span><\/p>\n

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

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

sin 30<\/span><\/span>o<\/span><\/span><\/sup> = v<\/span><\/span>oy<\/span><\/sub> \/ v<\/span><\/span>o<\/span><\/sub><\/p>\n

v<\/span><\/span>oy<\/span><\/sub> = v<\/span><\/span>o<\/span><\/sub> sin 30<\/span><\/span>o <\/span><\/span><\/sup>= (4)(sin 30<\/span><\/span>o<\/span><\/span><\/sup>) = (4)(0.5) = 2 <\/span><\/span>m\/s<\/span><\/span><\/p>\n

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

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

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

Vertical component of initial velocity (v<\/span><\/span>oy<\/span><\/sub>) = +2 <\/span><\/span>m\/s <\/span><\/span>(positive upward)<\/i><\/span><\/span><\/p>\n

Final velocity at maximum height (v<\/span><\/span>ty<\/span><\/sub>) = 0<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> The maximum height<\/span><\/span><\/p>\n

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

The maximum height :<\/span><\/span><\/p>\n

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

0<\/span><\/span>2<\/span><\/span><\/sup> = <\/span><\/span>2<\/span><\/span>2<\/span><\/span><\/sup> + 2 (-10) h<\/span><\/span><\/p>\n

0 <\/span><\/span>= 4<\/span><\/span> \u2013 20 h<\/span><\/span><\/p>\n

4 =<\/span><\/span> 20 h<\/span><\/span><\/p>\n

h = 4 \/ 20<\/span><\/span><\/p>\n

h = 0.2 meter<\/span><\/span><\/p>\n

The maximum height is 0.2 meter + 20 meter = 20.2 meter.<\/span><\/span><\/p>\n

<\/p>\n

[wpdm_package id=’528′]<\/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 maximum height 1. A kicked football leaves the ground at an angle \u03b8 = 60o with the horizontal has an initial speed of 10 m\/s. Calculate the maximum height! Acceleration of gravity is 10 m\/s2. Known : Angle (\u03b8) = 60o Initial speed (vo) = 10 m\/s … 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 maximum height 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-65","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\/65","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=65"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/65\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=65"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=65"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=65"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}