A tiger leaps horizontally from a 7.5-m-high rock with a speed of 3.0 m\/s. How far from the base of the rock will she land?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s solve it step-by-step:<\/p>\n\n\n\n First, find the time<\/strong> it takes the tiger to fall from the rock.<\/p>\n\n\n\n The vertical motion is free fall under gravity. We use the kinematic equation:<\/p>\n\n\n\n [ where:<\/p>\n\n\n\n Solving for ( t ):<\/p>\n\n\n\n [ Next, find the horizontal distance<\/strong>.<\/p>\n\n\n\n Since there\u2019s no air resistance and no horizontal acceleration, the horizontal distance ( x ) is:<\/p>\n\n\n\n [ where:<\/p>\n\n\n\n Substituting:<\/p>\n\n\n\n [ \u2705 Final Answer: The tiger will land approximately 3.72 meters away from the base of the rock.<\/strong><\/p>\n\n\n\n This problem is a classic projectile motion<\/strong> question where an object is launched horizontally. The key feature of horizontal launches is that the initial vertical velocity is zero<\/strong> because the tiger leaps straight out, not upwards or downwards.<\/p>\n\n\n\n The vertical motion is independent of the horizontal motion. Vertically, the only force acting on the tiger is gravity<\/strong>, which causes it to accelerate downwards at ( 9.8 \\, \\text{m\/s}^2 ). To find how long it takes the tiger to fall, we apply the formula for displacement under constant acceleration, ( y = \\frac{1}{2}gt^2 ). We solve this equation for time, and by substituting the given height (7.5 meters), we find the tiger falls for about 1.24 seconds.<\/p>\n\n\n\n Horizontally, there\u2019s no acceleration because we ignore air resistance. Thus, the horizontal velocity remains constant at ( 3.0 \\, \\text{m\/s} ) throughout the leap. Since velocity is constant, we can use the simple relationship ( x = v_x \\times t ) to find the horizontal distance traveled. Multiplying the tiger\u2019s constant horizontal speed by the time it\u2019s in the air gives the distance she covers horizontally before hitting the ground.<\/p>\n\n\n\n This separation of horizontal and vertical components makes projectile motion problems manageable: you treat them separately and combine the results at the end.<\/p>\n\n\n\n Thus, the tiger will land approximately 3.72 meters<\/strong> from the base of the rock.<\/p>\n\n\n\n A tiger leaps horizontally from a 7.5-m-high rock with a speed of 3.0 m\/s. How far from the base of the rock will she land? The correct answer and explanation is : Let’s solve it step-by-step: First, find the time it takes the tiger to fall from the rock. The vertical motion is free fall […]<\/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-208732","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208732","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=208732"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208732\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=208732"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=208732"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=208732"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n
y = \\frac{1}{2}gt^2
]<\/p>\n\n\n\n\n
t = \\sqrt{\\frac{2y}{g}}
]
[
t = \\sqrt{\\frac{2(7.5)}{9.8}}
]
[
t = \\sqrt{\\frac{15}{9.8}}
]
[
t = \\sqrt{1.5306}
]
[
t \\approx 1.24 \\, \\text{seconds}
]<\/p>\n\n\n\n
\n\n\n\n
x = v_x \\times t
]<\/p>\n\n\n\n\n
x = 3.0 \\times 1.24
]
[
x = 3.72 \\, \\text{meters}
]<\/p>\n\n\n\n
\n\n\n\n
\n\n\n\nDetailed Explanation (300+ words):<\/h3>\n\n\n\n
\n","protected":false},"excerpt":{"rendered":"