{"id":1047,"date":"2018-02-24T07:22:27","date_gmt":"2018-02-23T23:22:27","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1047"},"modified":"2018-02-24T07:22:27","modified_gmt":"2018-02-23T23:22:27","slug":"application-of-conservation-of-mechanical-energy-for-projectile-motion-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/application-of-conservation-of-mechanical-energy-for-projectile-motion-problems-and-solutions.htm","title":{"rendered":"Application of conservation of mechanical energy for projectile motion \u2013 problems and solutions","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>with the initial velocity of 10 m\/s. Ball’s mass<\/a> = 0.1 kg. Acceleration due to gravity<\/a> is 1<\/span><\/span>0 m\/s<\/span><\/span>2<\/span><\/span><\/sup>.<\/span><\/span> Determine (a) The gravitational potential energy<\/a> at the highest point (b) The highest point or the maximum height<\/span><\/span><\/p>\n

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

Mass (m) = 0.1 kg<\/span><\/span><\/p>\n

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

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

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

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

(a) The gravitational potential energy<\/b><\/span><\/span><\/p>\n

\"Application<\/p>\n

Calculate the horizontal component (v<\/span><\/span>ox<\/span><\/sub>) and the vertical component (v<\/span><\/span>oy<\/span><\/sub>) of initial velocity.<\/span><\/span><\/p>\n

\"Application\"Applicationv<\/span><\/span>ox <\/span><\/sub>= v<\/span><\/span>o<\/span><\/sub> cos <\/span><\/span>\u03b8<\/span><\/span> = (10)(cos 30<\/span><\/span>o<\/span><\/span><\/sup>) = (10)(<\/span><\/span>0.5\u221a3) = 5\u221a3 m\/s<\/span><\/span><\/p>\n

v<\/span><\/span>oy <\/span><\/sub>= v<\/span><\/span>o<\/span><\/span><\/sub> sin <\/span><\/span>\u03b8 <\/span><\/span> = (10)(sin 30<\/span><\/span>o<\/span><\/span><\/sup>) = (10)(<\/span><\/span>0.5) = 5 m\/s<\/span><\/span><\/p>\n

The initial mechanical energy<\/b><\/span><\/span><\/p>\n

The initial mechanical energy<\/a> (ME<\/span><\/span>o<\/span><\/sub>) = kinetic energy<\/a> (KE)<\/span><\/span><\/p>\n

ME<\/span><\/span>o <\/span><\/sub>= KE = \u00bd m v<\/span><\/span>o<\/span><\/sub>2<\/span><\/span><\/sup> = \u00bd (0.1)(10)<\/span><\/span>2<\/span><\/span><\/sup> = \u00bd (0.1)(100) = \u00bd (10) = 5 Joule <\/span><\/span><\/p>\n

The final mechanical energy<\/b><\/span><\/span><\/p>\n

Kinetic energy at the highest point :<\/span><\/span><\/p>\n

KE = \u00bd m v<\/span><\/span>ox<\/span><\/sub>2<\/span><\/span><\/sup> = \u00bd (0.1)(5\u221a3)<\/span><\/span>2 <\/span><\/span><\/sup>= \u00bd (0.1)((25)(3)) = \u00bd (0.1)(75) = 3.75 Joule <\/span><\/span><\/p>\n

Principle of conservation of mechanical energy <\/b><\/span><\/span><\/p>\n

The initial mechanical energy (ME<\/span><\/span>o<\/span><\/sub>) = the final mechanical energy (ME<\/span><\/span>t<\/span><\/sub>)<\/span><\/span><\/p>\n

KE = PE + KE<\/span><\/span><\/p>\n

5 = EP + 3.75<\/span><\/span><\/p>\n

PE = 5 \u2013 3.75 = 1.25 Joule<\/span><\/span><\/p>\n

The gravitational potential energy at the highest point is 1.25 Joule.<\/span><\/span><\/p>\n

(b) The highest point or the maximum height<\/b><\/span><\/span><\/p>\n

PE = m g h<\/span><\/span><\/p>\n

1.25 = (0.1)(10) h<\/span><\/span><\/p>\n

1.25 = h<\/span><\/span><\/p>\n

The maximum height is 1.25 meters.<\/span><\/span><\/p>\n

<\/p>\n

2. <\/span><\/span>A 0.1-kg ball projected horizontally with initial velocity v<\/span><\/span>o<\/span><\/sub> = 10 m\/s from a building 10 meter high. <\/span><\/span>Acceleration due to gravity is 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup>. Determine ball’s kinetic energy when it hits the ground.<\/span><\/span><\/p>\n

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

Mass (m) = 0.1 kg<\/span><\/span><\/p>\n

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

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

The change in height (h) = 10 \u2013 2 = 8 m<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> kinetic energy at 2 meters above the ground<\/span><\/span><\/p>\n

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

The gravitational potential energy (PE) = m g h = (0.1)(10)(10) = 10 Joule<\/span><\/span><\/p>\n

The initial kinetic energy (KE)= \u00bd m v<\/span><\/span>o<\/span><\/span><\/sub>2 <\/span><\/span><\/sup>= \u00bd (0.1)(10)<\/span><\/span>2<\/span><\/span><\/sup> = \u00bd (0.1)(100) = \u00bd (10) = 5 Joule<\/span><\/span><\/p>\n

The final kinetic energy = the initial gravitational potential energy + the initial kinetic energy = 10 + 5 = 15 Joule<\/span><\/span><\/p>\n

[wpdm_package id=’1173′]<\/p>\n

    \n
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  2. Work-kinetic energy problems and solutions<\/a><\/li>\n
  3. Work-mechanical energy principle problems and solutions<\/a><\/li>\n
  4. Gravitational potential energy problems and solutions<\/a><\/li>\n
  5. The potential energy of elastic spring problems and solutions<\/a><\/li>\n
  6. Power problems and solutions<\/a><\/li>\n
  7. Application of conservation of mechanical energy for free fall motion<\/a><\/li>\n
  8. Application of conservation of mechanical energy for up and down motion in free fall motion<\/a><\/li>\n
  9. Application of conservation of mechanical energy for motion on a curved surface<\/a><\/li>\n
  10. Application of conservation of mechanical energy for motion on an inclined plane<\/a><\/li>\n
  11. Application of conservation of mechanical energy for projectile motion<\/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 with the initial velocity of 10 m\/s. Ball’s mass = 0.1 kg. Acceleration due to gravity is 10 m\/s2. Determine (a) The gravitational potential energy at the highest point (b) The highest point or the maximum height Known : Mass (m) = … 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":"Application of conservation of mechanical energy for projectile motion \u2013 problems and solutions","_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-1047","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\/1047","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=1047"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1047\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1047"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1047"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1047"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}