{"id":71,"date":"2018-01-14T05:26:53","date_gmt":"2018-01-13T21:26:53","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=71"},"modified":"2018-01-14T05:26:53","modified_gmt":"2018-01-13T21:26:53","slug":"solving-projectile-motion-problems-determine-time-interval","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/solving-projectile-motion-problems-determine-time-interval.htm","title":{"rendered":"Determine time interval of projectile motion","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Solved problems in projectile motion<\/a> – determine the time interval<\/a><\/p>\n

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 10 m\/s. Calculate the time interval to reach 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) = 3<\/span><\/span>0<\/span><\/span>o<\/span><\/span><\/sup><\/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> Time interval to reach the maximum height<\/a><\/span><\/span><\/p>\n

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

\"SolvingVertical 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) =<\/span><\/span> 5 <\/span><\/span>m\/s<\/span><\/span><\/p>\n

Time interval to reach maximum height is determined by the vertical motion<\/a> equation. 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> 5 <\/span><\/span>m\/s<\/span><\/span> (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

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

Wanted :<\/u><\/span><\/span> time interval (t)<\/span><\/span><\/p>\n

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

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

0 = 5 + (-10)t<\/span><\/span><\/p>\n

0 = 5 \u2013 10 t<\/span><\/span><\/p>\n

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

t = 5\/10 = 0.5 s<\/span><\/span><\/p>\n

<\/p>\n

2. A body is projected upward at angle of 30<\/span><\/span>o <\/span><\/span><\/sup>to <\/span><\/span>the horizontal with an initial speed of 30 m\/s. Calculate time of flight! 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) = 3<\/span><\/span>0<\/span><\/span>o<\/span><\/span><\/sup><\/p>\n

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = 8 m\/s<\/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> Time interval before body hits the ground<\/span><\/span><\/p>\n

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

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

v<\/span><\/span>oy <\/span><\/sub>= v<\/span><\/span>o<\/span><\/sub> sin \u03b8 = (8 m\/s)(sin 30<\/span><\/span>o<\/span><\/span><\/sup>) = (8 m\/s)(0.5) =<\/span><\/span> 4 <\/span><\/span>m\/s<\/span><\/span><\/p>\n

We first calculate time interval to reach the maximum height using equation of vertical motion.<\/span><\/span><\/p>\n

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

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

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = <\/span><\/span>4 <\/span><\/span>m\/s<\/span><\/span> (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

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

Wanted :<\/u><\/span><\/span> Time interval (t)<\/span><\/span><\/p>\n

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

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

0 = 4 + (-10)t<\/span><\/span><\/p>\n

0 = 4 \u2013 10 t<\/span><\/span><\/p>\n

4 = 10 t<\/span><\/span><\/p>\n

t = 4\/10 = 0,4 s<\/span><\/span><\/p>\n

Time interval to reach the maximum height is 0.4 s. <\/span><\/span><\/p>\n

Time in air is 2 x 0.4 s = 0.8 s.<\/span><\/span><\/p>\n

<\/p>\n

3. A body is projected upward at an angle of 30<\/span><\/span>o <\/span><\/span><\/sup>with the horizontal from a building 10 meters high. Its initial speed is 40 m\/s. How long does it take the body to reach 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 (\u03b8) = 3<\/span><\/span>0<\/span><\/span>o<\/span><\/span><\/sup><\/p>\n

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

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = 40 m\/s<\/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> Time in air (t)<\/span><\/span><\/p>\n

Solution :<\/u><\/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 = (40 m\/s)(sin 30<\/span><\/span>o<\/span><\/span><\/sup>) = (40 m\/s)(0.5) = 20<\/span><\/span> m\/s<\/span><\/span><\/p>\n

We first calculate time interval to reach the maximum height using equation of vertical motion.<\/span><\/span><\/p>\n

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

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

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = <\/span><\/span>20 <\/span><\/span>m\/s<\/span><\/span> (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

Final velocity at peak (v<\/span><\/span>t<\/span><\/sub>) = 0 <\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Time interval (t)<\/span><\/span><\/p>\n

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

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

0 = 20 + (-10)t<\/span><\/span><\/p>\n

0 = 20 \u2013 10 t<\/span><\/span><\/p>\n

20 = 10 t<\/span><\/span><\/p>\n

t = 20\/10 = 2 seconds<\/span><\/span><\/p>\n

Time in air = 2 x 2 seconds = 4 seconds.<\/span><\/span><\/p>\n

The object is 10 meters above the ground. 4 seconds is the time interval to reach a place that parallels to the initial position. The ball is still moving downward.<\/span><\/span><\/span><\/p>\n

The time interval to reach the ground is calculated using the equation of free fall motion<\/a>.\u00a0 <\/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

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

Wanted :<\/u><\/span><\/span> Time interval (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

Time interval = 1.4 seconds.<\/span><\/span><\/p>\n

Total time interval = 4 seconds + 1.4 seconds = 5.4 seconds.<\/span><\/span><\/p>\n

<\/p>\n

4. <\/span><\/span>A small ball projected horizontally with initial velocity v<\/span><\/span>o<\/span><\/sub> = 15 m\/s from a building 5 meters high. Calculate time in the air<\/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) = 5 meters<\/span><\/span><\/p>\n

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = 15 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> Time in the air (t)<\/span><\/span><\/p>\n

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

\"SolvingTime in the air is calculated using the equation of freely falling motion. <\/span><\/span><\/p>\n

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

High (h) = 5 meters<\/span><\/span><\/p>\n

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

Wanted :<\/u><\/span><\/span> Time interval (t)<\/span><\/span><\/p>\n

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

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

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

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

t<\/span><\/span>2 <\/span><\/span><\/sup>= 5\/5 = 1<\/span><\/span><\/p>\n

t = <\/span><\/span>\u221a1 = 1 second<\/span><\/span><\/p>\n

<\/p>\n

[wpdm_package id=’531′]<\/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 time interval 1. A kicked football leaves the ground at an angle \u03b8 = 30o to the horizontal with an initial speed of 10 m\/s. Calculate the time interval to reach the maximum height! Acceleration of gravity is 10 m\/s2. Known : Angle (\u03b8) = 30o Initial … 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 time interval 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-71","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\/71","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=71"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/71\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=71"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=71"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=71"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}