A projectile is shot from the edge of a cliff 115 m above ground level with an initial speed of 65.0 m\/s at an angle of 35.0 \u2218 with the horizontal.<\/p>\n\n\n\n
A). Determine the time taken by the projectile to hit point P at ground level. t= s<\/p>\n\n\n\n
B). Determine the distance X of point P from the base of the vertical cliff. Deltax= m<\/p>\n\n\n\n
C). At the instant just before the projectile hits point P, determine the horizontal and the vertical components of its velocity.<\/p>\n\n\n\n
Enter your answers numerically separated by a comma. Vx, Vy= ,<\/em><\/strong> m\/s<\/p>\n\n\n\n D). At the instant just before the projectile hits point P, determine the magnitude of the velocity. v= m\/s<\/p>\n\n\n\n E). At the instant just before the projectile hits point P, determine the angle made by the velocity vector with the horizontal. Theta= degrees below the horizon.<\/p>\n\n\n\n The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To solve this projectile motion problem, let’s break it down step by step. Given:<\/p>\n\n\n\n We first determine the total time of flight using the vertical motion equation:<\/p>\n\n\n\n [ where [ Solving this quadratic equation:<\/p>\n\n\n\n [ Using the quadratic formula:<\/p>\n\n\n\n [ [ [ [ Taking the positive root:<\/p>\n\n\n\n [ The horizontal component of velocity remains constant:<\/p>\n\n\n\n [ [ The horizontal velocity remains constant:<\/p>\n\n\n\n [ For vertical velocity at impact:<\/p>\n\n\n\n [ [ (negative sign indicates downward direction) [ [ [ [ [ Final Answers: A projectile is shot from the edge of a cliff 115 m above ground level with an initial speed of 65.0 m\/s at an angle of 35.0 \u2218 with the horizontal. A). Determine the time taken by the projectile to hit point P at ground level. t= s B). Determine the distance X of point […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"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":"","ast-disable-related-posts":"","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":[25],"tags":[],"class_list":["post-205709","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/205709","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=205709"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/205709\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=205709"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=205709"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=205709"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
A) Time taken to hit the ground<\/strong><\/h3>\n\n\n\n
y = y_0 + v_{0y} t – \\frac{1}{2} g t^2
]<\/p>\n\n\n\n
( v_{0y} = v_0 \\sin\\theta_0 = 65.0 \\sin 35^\\circ = 37.24 ) m\/s.
At impact, ( y = 0 ):<\/p>\n\n\n\n
0 = 115 + (37.24)t – \\frac{1}{2} (9.81)t^2
]<\/p>\n\n\n\n
4.905t^2 – 37.24t – 115 = 0
]<\/p>\n\n\n\n
t = \\frac{-(-37.24) \\pm \\sqrt{(-37.24)^2 – 4(4.905)(-115)}}{2(4.905)}
]<\/p>\n\n\n\n
t = \\frac{37.24 \\pm \\sqrt{1387.4 + 2254.3}}{9.81}
]<\/p>\n\n\n\n
t = \\frac{37.24 \\pm \\sqrt{3641.7}}{9.81}
]<\/p>\n\n\n\n
t = \\frac{37.24 \\pm 60.36}{9.81}
]<\/p>\n\n\n\n
t = \\frac{37.24 + 60.36}{9.81} = \\frac{97.6}{9.81} = 9.95 \\text{ s}
]<\/p>\n\n\n\nB) Horizontal Distance (Range)<\/strong><\/h3>\n\n\n\n
v_{0x} = v_0 \\cos\\theta_0 = 65.0 \\cos 35^\\circ = 53.27 \\text{ m\/s}
]<\/p>\n\n\n\n
\\Delta x = v_{0x} \\cdot t = 53.27 \\times 9.95 = 530.1 \\text{ m}
]<\/p>\n\n\n\nC) Velocity Components Just Before Impact<\/strong><\/h3>\n\n\n\n
V_x = 53.27 \\text{ m\/s}
]<\/p>\n\n\n\n
V_y = v_{0y} – g t = 37.24 – (9.81 \\times 9.95)
]<\/p>\n\n\n\n
V_y = 37.24 – 97.65 = -60.41 \\text{ m\/s}
]<\/p>\n\n\n\n
Thus, ( V_x, V_y = 53.27, -60.41 ) m\/s.<\/p>\n\n\n\nD) Magnitude of Velocity<\/strong><\/h3>\n\n\n\n
V = \\sqrt{V_x^2 + V_y^2}
]<\/p>\n\n\n\n
V = \\sqrt{(53.27)^2 + (-60.41)^2}
]<\/p>\n\n\n\n
V = \\sqrt{2837.7 + 3649.1} = \\sqrt{6486.8} = 80.55 \\text{ m\/s}
]<\/p>\n\n\n\nE) Angle with Horizontal<\/strong><\/h3>\n\n\n\n
\\theta = \\tan^{-1} \\left( \\frac{|V_y|}{V_x} \\right) = \\tan^{-1} \\left( \\frac{60.41}{53.27} \\right)
]<\/p>\n\n\n\n
\\theta = \\tan^{-1} (1.134) = 48.2^\\circ
]<\/p>\n\n\n\n
A) t = 9.95 s<\/strong>
B) \u0394x = 530.1 m<\/strong>
C) ( V_x, V_y = 53.27, -60.41 ) m\/s<\/strong>
D) ( V = 80.55 ) m\/s<\/strong>
E) ( \\theta = 48.2^\\circ ) below the horizon<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"