concept, formula, and derivation<\/strong><\/span><\/p>\n In this article, we will learn about horizontal projectile motion. In this case, the projectile is launched or fired parallel to horizontal. So, it starts with a horizontal initial velocity, some height ‘h’ and no vertical velocity. I am assuming that you know about the basic concepts of projectile motion<\/a>.<\/p>\n Projectile motion is a form of motion in which an object or particle (called a projectile) is thrown with some initial velocity near the earth’s surface, and it moves along a curved path under the action of gravity alone.<\/p><\/blockquote>\n For the derivation of various formulas for horizontal projectile motion, consider the figure given below,<\/p>\n Suppose a body is thrown horizontally from a point \\(O\\) with velocity \\(u\\). Point \\(O\\) is at height \\(h\\) above the ground. After a time t suppose the body reaches point \\(P\\left( x,y \\right)\\) then, Along the horizontal axis, It is the total time for which the projectile remains in flight (from \\(O\\) to \\(A\\)). Let \\(T\\) be the time of flight. For vertical downward motion of the body we use \\begin{align*} It is the horizontal distance covered by projectile during the time of flight. It is equal to \\(OA=R\\). So, \\begin{align*} Horizontally launched projectile concept, formula, and derivation In this article, we will learn about horizontal projectile motion. In this case, the projectile is launched or fired parallel to horizontal. So, it starts with a horizontal initial velocity, some height ‘h’ and no vertical velocity. I am assuming that you know about the basic concepts of […]<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[14],"tags":[],"class_list":["post-5475","post","type-post","status-publish","format-standard","hentry","category-physics"],"yoast_head":"\n
\n![\"horizontal](\"https:\/\/physicscatalyst.com\/article\/wp-content\/uploads\/2019\/10\/projectile-motion-horizontal.png\")
<\/a>
\nNow the body has two simultaneously independent motions.
\n(i) motion with uniform horizontal velocity \\(u\\).
\n(ii) vertically downward accelerated motion with constant acceleration \\(g\\).
\nA body moves along path \\(OPA\\) under the influence of these two independent motions.<\/p>\nTrajectory of the projectile<\/h3>\n
\nAlong horizontal axis at
\n\\(u_x = u\\) (since motion is with uniform horizontal velocity)
\n\\(a_x = 0\\)
\n\\(s_x=x\\)
\n\\(distance = speed \\times time\\) or, \\(x = u\\times \\,t\\)
\n\\[t = \\frac{x}{u} \\tag{1}\\]
\nAlong vertical axis ,
\n\\(u_y = 0 \\) at time \\(t=0\\)
\n\\(a_y = g\\)
\n\\(s_y = y\\)
\nBy second equation of motion,
\n\\begin{align*}
\n&s_y = u_yt + \\frac{1}{2} a_yt^2\\\\
\n&\\text{Using equation (1) } \\text{and putting} u_y = 0 \\text{at time} t=0 \\\\
\n&y = \\text{0 }+ \\frac{1}{2} g \\left( \\frac{x}{u} \\right) ^2\\\\
\n&y = \\frac{1}{2} g \\frac{x^2}{u^2}\\\\
\n&y = \\frac{g x^2}{\\text{2 }u^2}
\n\\end{align*}
\nThis is the equation of parabola. So, the trajectory of the projectile fired parallel to the horizontal is a parabola.<\/p>\nThe velocity of the projectile at any time<\/h3>\n
\n\\(a_x = 0\\)
\nso, velocity remains constant and velocity at \\(A\\) along horizontal will also be \\(u\\).
\nAlong vertical,
\n\\(u_y = 0\\)
\n\\(a_y = g\\)
\nBy first equation of motion
\n\\(v_y = u_y + a_yt\\)
\n\\(v_y = 0 + gt\\)
\n\\(v_y = gt\\)
\nMagnitude of resultant velocity at any point \\(P\\)
\n\\begin{align*}
\n&v^2 = v_x^2 + v_y^2 \\\\
\n&v= \\sqrt{u_x^2 + g^2t^2}\\end{align*}
\nNow \\(\\beta\\) is the angle which resultant velocity makes with the horizontal, then
\n\\[\\tan \\beta =\\frac{ v_y^2}{v_x^2\\,}=\\frac{gt}{u}\\]
\nor,
\n\\[\\beta =\\tan ^{-1}\\left( \\frac{gt}{u} \\right) \\]<\/p>\nTime of flight<\/h3>\n
\n&s_y = u_gt + \\frac{1}{2} a_yt^2\\\\
\n&\\text{or,}\\\\
\n&h=0\\times T +\\frac{1}{2}gT^2\\\\
\n&\\text{or,}\\\\
\n&T=\\sqrt{\\frac{2h}{g}}
\n\\end{align*}<\/p>\nHorizontal Range<\/h3>\n
\nR&=\\text{Horizontal velocity} \\times \\text{Time of flight}\\\\
\n&=u\\times T = u\\sqrt{\\frac{2h}{g}}
\n\\end{align*}So,\\[R=u\\sqrt{\\frac{2h}{g}}\\]<\/p>\n
\nRelated Articles and links<\/h4>\n
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