{"id":2026,"date":"2018-04-21T14:27:24","date_gmt":"2018-04-21T06:27:24","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2026"},"modified":"2023-08-16T21:59:52","modified_gmt":"2023-08-16T21:59:52","slug":"projectile-motion-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/projectile-motion-problems-and-solutions.htm","title":{"rendered":"Projectile motion \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

27 Projectile motion \u2013 problems and solutions<\/span><\/p>\n

1. A bullet fired a<\/span>t <\/span>an angle \u03b8 = 60<\/span>o <\/span><\/sup>with a velocity of <\/span>20 m\/s. <\/span>Acceleration due to gravity<\/a> is <\/span>10 m\/s2<\/sup>. What is the time interval to reach the maximum height?<\/span><\/p>\n

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

The initial velocity of bullet (vo<\/sub>) = 20 m\/s<\/span><\/p>\n

Angle (\u03b8) = 60o<\/sup>C<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m s\u20132<\/sup><\/span><\/p>\n

Wanted :<\/u> The time interval to reach the maximum height<\/a><\/span><\/p>\n

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

The initial velocity at the horizontal direction (x axis) :<\/span><\/p>\n

vox<\/sub> = vo<\/sub> cos 60o<\/sup> = (20)(0.5) = 10 m\/s<\/span><\/p>\n

The initial velocity at the vertical direction (y axis) :<\/span><\/p>\n

voy<\/sub> = vo sin 60o<\/sup> = (20)(0.5\u221a3) = 10\u221a3 m\/s<\/span><\/p>\n

The time interval to reach the maximum height, calculated using this equation :<\/span><\/p>\n

vty<\/sub> = voy<\/sub> + g t<\/span><\/p>\n

vty<\/sub> = the final velocity in the vertical direction = the final velocity at the highest point = 0 m\/s<\/span><\/p>\n

voy <\/sub>= the initial velocity at the horizontal direction = 10\u221a3 m\/s <\/span><\/p>\n

g = acceleration due to gravity = 10 m\/s2<\/sup><\/span><\/p>\n

t = time interval<\/span><\/p>\n

The time interval :<\/span><\/p>\n

vty<\/sub> = voy<\/sub> + g t<\/span><\/p>\n

0 = 10\u221a3 \u2013 10 t<\/span><\/p>\n

10\u221a3 = 10 t<\/span><\/p>\n

t = 10\u221a3 \/ 10 <\/span><\/p>\n

t = \u221a3 seconds<\/span><\/p>\n

2. An object projected at an angle. The height of the object is the same when the time interval = 1 second and 3 seconds. What is the time interval the object in air.<\/span><\/p>\n

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

\"Projectile<\/span><\/p>\n

The object in the air for 4 seconds.<\/span><\/p>\n

3. An aircraft is moving <\/span>horizontally <\/span>with a speed of 50 m\/s. <\/span>At the height of 2 km, an object is dropped from the aircraft. Acceleration due to gravity = 10 m\/<\/span>s2, what<\/span> is the time interval before the object hits the ground. <\/span><\/span><\/p>\n

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

Height = 2 km = 2000 meters\"Projectile<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s2<\/sup><\/span><\/p>\n

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

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

h = 1\/2 g t2<\/sup><\/span><\/p>\n

2000 = 1\/2 (10) t2<\/sup><\/span><\/p>\n

2000 = 5 t2<\/sup><\/span><\/p>\n

t2 <\/sup>= 2000\/5 = 400<\/span><\/p>\n

t = \u221a400 = 20 seconds<\/span><\/p>\n

4. A kicked football leaves the ground at an angle \u03b8 = 45o <\/sup>with the horizontal has an initial speed of 25 m\/s. Determine the distance of X. Acceleration due to gravity is 10 m\/s2<\/sup>.<\/span><\/p>\n

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

Initial speed (vo<\/sub>) = 25 m\/s<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s2<\/sup><\/span><\/p>\n

Angle (\u03b8) = 45o<\/sup><\/span><\/p>\n

Wanted :<\/u> X<\/span><\/p>\n

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

The horizontal component of the initial velocity :<\/span><\/p>\n

vox<\/sub> = vo<\/sub> cos \u03b8 = (25 m\/s)(cos 45o<\/sup>) = (25 m\/s)(0.5\u221a2) = 12.5\u221a2 m\/s<\/span><\/p>\n

The vertical component of the initial velocity :<\/span><\/p>\n

voy <\/sub>= vo<\/sub> sin \u03b8 = (25 m\/s)(sin 45o<\/sup>) = (25 m\/s)(0.5\u221a2) = 12.5\u221a2 m\/s<\/span><\/p>\n

Projectile motion could be understood by analyzing the horizontal and vertical component of the motion separately. The x motion occurs at constant velocity and the y motion occurs at constant acceleration of gravity. <\/span><\/p>\n

Time in the air<\/b><\/a> <\/b>(t) :<\/b><\/span><\/p>\n

The time in air calculated with the equation of the upward vertical motion.<\/span><\/p>\n

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

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

The initial velocity (vo<\/sub>) = 12.5\u221a2 m\/s (upward direction, positive)<\/span><\/p>\n

Acceleration due to gravity (g) = -10 m\/s2 <\/sup>(downward direction, negative)<\/span><\/p>\n

Height (h) = 0 <\/span><\/p>\n

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

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

h = vo<\/sub> t + 1\/2 g t2<\/sup><\/span><\/p>\n

0 = (12.5\u221a2) t + 1\/2 (-10) t2<\/sup><\/span><\/p>\n

0 = 12.5\u221a2 t – 5 t2<\/sup><\/span><\/p>\n

12.5\u221a2 t = 5 t2<\/sup><\/span><\/p>\n

12.5\u221a2 = 5 t<\/span><\/p>\n

t = 12.5\u221a2 \/ 5 <\/span><\/p>\n

t = 2.5\u221a2 seconds<\/span><\/p>\n

The horizontal distance <\/b>(X) :<\/b><\/span><\/p>\n

Calculated using the equation of the uniform linear motion<\/a> with constant velocity.<\/span><\/p>\n

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

Velocity (v) = 12.5\u221a2 m\/s<\/span><\/p>\n

Time interval (t) = 2.5\u221a2 seconds<\/span><\/p>\n

Wanted :<\/u> Distance<\/span><\/p>\n

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

d = v t = (12.5\u221a2)(2.5\u221a2) = (12.5)(2.5)(2) = 62.5 meters<\/span><\/p>\n

5. An object projected upward at an angle \u03b8 = 30o <\/sup>with the horizontal has an initial speed of 20 m\/s. Acceleration due to gravity is 10 m\/s2<\/sup>. Determine the maximum height.<\/span><\/p>\n

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

The initial velocity (vo<\/sub>) = 20 m\/s\"Projectile<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s2<\/sup><\/span><\/p>\n

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

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

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

First, find the vertical component of the initial velocity (voy<\/sub>) :<\/span><\/p>\n

voy<\/sub> = vo<\/sub> sin 30o <\/sup>= (20)(sin 30o<\/sup>) = (20)(0.5) = 10 m\/s<\/span><\/p>\n

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

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

Acceleration due to gravity (g) = -10 m\/s2 <\/sup>(<\/i>downward <\/i>direction, negative<\/i>)<\/i><\/span><\/p>\n

The vertical component of the initial velocity (voy<\/sub>) = 10 m\/s (<\/i>upward direction, positive<\/i>)<\/i><\/span><\/p>\n

Velocity at the maximum height (vty<\/sub>) = 0<\/span><\/p>\n

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

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

vt<\/sub>2<\/sup> = vo<\/sub>2<\/sup> + 2 g h<\/span><\/p>\n

02<\/sup> = 102<\/sup> + 2 (-10) h<\/span><\/p>\n

0 = 100 \u2013 20 h<\/span><\/p>\n

100 = 20 h<\/span><\/p>\n

h = 100\/20<\/span><\/p>\n

h = 5 meters<\/span><\/p>\n

The maximum height is 5 meters.<\/span><\/p>\n

6. An object is thrown at a certain elevation angle. The height of the object same after 1 second and 3 seconds. Determine time in air.<\/span><\/span><\/p>\n

A. 3.6 s\"Projectile<\/span><\/p>\n

B. 4.0 s<\/span><\/p>\n

C. 5.6 s<\/span><\/p>\n

D. 6.4 s<\/span><\/p>\n

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

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

The correct answer is B.<\/span><\/p>\n

7. An aircraft is moving horizontally with the speed of 50 m\/s. When the aircraft at the height of 2 km, an object free fall from the aircraft. Determine the type of the motion.<\/span><\/p>\n

A. Free fall motion\"Projectile<\/span><\/p>\n

B. Floating motion<\/span><\/p>\n

C. Horizontal motion<\/span><\/p>\n

D. Projectile motion<\/span><\/p>\n

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

The object is dropped from the moving plane because it has the same speed as the plane’s speed, that is 50 m\/s. Movement of objects is not like free fall motion but parabolic motion. The case is the same as you are dropping objects from inside a moving car.<\/span><\/span><\/p>\n

The correct answer is D.<\/span><\/p>\n

8. A ball is thrown horizontally at 15 m\/s from a cliff 60 meters high. How long does it take to hit the ground?<\/span>
\nSolution: Using \\( h = \\frac{1}{2} g t^2 \\), we find the time is \\( t = \\sqrt{\\frac{2h}{g}} \\approx 3.5\\ \\text{s} \\).<\/span><\/p>\n

9. A projectile is fired at an angle of 30\u00b0 above the horizontal with an initial speed of 20 m\/s. What is the maximum height reached?<\/span>
\nSolution: Using \\( h = \\frac{v^2 \\sin^2 \\theta}{2g} \\), the maximum height is \\( h \\approx 10.2\\ \\text{m} \\).<\/span><\/p>\n

10. A stone is thrown horizontally at 10 m\/s from a tower 80 meters tall. Find the horizontal distance it travels before hitting the ground.<\/span>
\nSolution: Using the time found in a similar way to Problem 1, the horizontal distance is \\( d = vt \\approx 40\\ \\text{m} \\).<\/span><\/p>\n

11. A cannonball is fired at 40 m\/s at an angle of 45\u00b0. Find the time of flight.<\/span>
\nSolution: Using \\( t = \\frac{2v \\sin \\theta}{g} \\), the time of flight is \\( t \\approx 5.8\\ \\text{s} \\).<\/span><\/p>\n

12.\u00a0 A baseball is thrown at an angle of 60\u00b0 with a velocity of 12 m\/s. Find the horizontal range.<\/span>
\nSolution: Using \\( R = \\frac{v^2 \\sin 2\\theta}{g} \\), the range is \\( R \\approx 14.0\\ \\text{m} \\).<\/span><\/p>\n

13. A projectile is launched with an initial velocity of 50 m\/s at 37\u00b0 above the horizontal. What is the vertical velocity component?<\/span>
\nSolution: The vertical component is \\( v_y = v \\sin \\theta \\approx 30\\ \\text{m\/s} \\).<\/span><\/p>\n

14. A projectile is launched horizontally at 20 m\/s from a height of 100 meters. What is the vertical velocity just before it hits the ground?<\/span>
\nSolution: Using \\( v_y = \\sqrt{2gh} \\), the vertical velocity is \\( v_y \\approx 44.7\\ \\text{m\/s} \\).<\/span><\/p>\n

15. A rock is thrown at an angle of 25\u00b0 above the horizontal with an initial speed of 15 m\/s. What are the horizontal and vertical components of the velocity?<\/span>
\nSolution: The horizontal component is \\( v_x = v \\cos \\theta \\approx 13.4\\ \\text{m\/s} \\), and the vertical component is \\( v_y \\approx 6.4\\ \\text{m\/s} \\).<\/span><\/p>\n

16. A soccer ball is kicked with an initial speed of 30 m\/s at an angle of 40\u00b0 above the horizontal. What is its horizontal velocity component?<\/span>
\nSolution: The horizontal component is \\( v_x = v \\cos \\theta \\approx 22.9\\ \\text{m\/s} \\).<\/span><\/p>\n

17. A golf ball is hit with an initial speed of 70 m\/s at an angle of 20\u00b0. What is the time of flight?<\/span>
\nSolution: Using the time of flight equation, the time is \\( t \\approx 4.9\\ \\text{s} \\).<\/span><\/p>\n

18. A projectile is fired from the ground with a speed of 25 m\/s at 53\u00b0 above the horizontal. What is its initial vertical velocity component?<\/span>
\nSolution: The vertical component is \\( v_y = v \\sin \\theta \\approx 20\\ \\text{m\/s} \\).<\/span><\/p>\n

19. A baseball is thrown with an initial speed of 20 m\/s at an angle of 50\u00b0. What is the maximum height?<\/span>
\nSolution: Using the equation for maximum height, the height is \\( h \\approx 15.3\\ \\text{m} \\).<\/span><\/p>\n

20. A bullet is fired horizontally with a velocity of 200 m\/s from a height of 10 meters. How long does it take to hit the ground?<\/span>
\nSolution: Using the time equation, the time is \\( t \\approx 1.4\\ \\text{s} \\).<\/span><\/p>\n

21. A cannonball is fired at 45 m\/s at an angle of 30\u00b0. Find the range.<\/span>
\nSolution: Using the range equation, the range is \\( R \\approx 88.2\\ \\text{m} \\).<\/span><\/p>\n

22. A basketball is thrown at an angle of 75\u00b0 with a velocity of 10 m\/s. Find the horizontal range.<\/span>
\nSolution: Using the range equation, the range is \\( R \\approx 5.3\\ \\text{m} \\).<\/span><\/p>\n

23. A projectile is launched with an initial velocity of 30 m\/s at 22\u00b0 above the horizontal. What is the vertical velocity just before it hits the ground?<\/span>
\nSolution: Using the vertical velocity equation, the vertical velocity is \\( v_y \\approx 11.4\\ \\text{m\/s} \\).<\/span><\/p>\n

24. A rock is thrown horizontally at 8 m\/s from a tower 40 meters tall. What is the horizontal distance it travels?<\/span>
\nSolution: Using the horizontal distance equation, the distance is \\( d \\approx 16\\ \\text{m} \\).<\/span><\/p>\n

25. A soccer ball is kicked with an initial speed of 12 m\/s at an angle of 30\u00b0 above the horizontal. What are the horizontal and vertical components of the velocity?<\/span>
\nSolution: The horizontal component is \\( v_x \\approx 10.4\\ \\text{m\/s} \\), and the vertical component is \\( v_y \\approx 6\\ \\text{m\/s} \\).<\/span><\/p>\n

26. A golf ball is hit with an initial speed of 50 m\/s at an angle of 15\u00b0. What is the time of flight?<\/span>
\nSolution: Using the time of flight equation, the time is \\( t \\approx 2.6\\ \\text{s} \\).<\/span><\/p>\n

27. A projectile is fired from the ground with a speed of 40 m\/s at 60\u00b0 above the horizontal. What is its initial horizontal velocity component?<\/span>
\nSolution: The horizontal component is \\( v_x = v \\cos \\theta \\approx 20\\ \\text{m\/s} \\).<\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

27 Projectile motion \u2013 problems and solutions 1. A bullet fired at an angle \u03b8 = 60o with a velocity of 20 m\/s. Acceleration due to gravity is 10 m\/s2. What is the time interval to reach the maximum height? Known : The initial velocity of bullet (vo) = 20 m\/s Angle (\u03b8) = 60oC … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"closed","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":"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-2026","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\/2026","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=2026"}],"version-history":[{"count":4,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2026\/revisions"}],"predecessor-version":[{"id":8959,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2026\/revisions\/8959"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=2026"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=2026"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=2026"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}