A grasshopper hops along a level road. On each hop, the grasshopper launches itself at angle \u03b80 = 45\u03bf and achieves a range R = 0.80 m. What is the average horizontal speed of the grasshopper as it hops along the road? Assume that the time spent on the ground between hops in negligible and achieves a range. What is the average horizontal speed of the grasshopper as it hops along the road? Assume that the time spent on the ground between hops is negligible.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To find the average horizontal speed of the grasshopper, we must first determine the time of flight for each hop and the horizontal component of velocity during the hop.<\/p>\n\n\n\n The range ( R ) of a projectile is given by: Given ( R = 0.80 \\, \\text{m} ), substituting ( \\sin 2\\theta_0 = \\sin 90^\\circ = 1 ): The time of flight ( T ) for a projectile is given by: The horizontal velocity of the grasshopper during each hop is: Since the time spent on the ground is negligible, the average horizontal speed<\/strong> is equal to the horizontal velocity during each hop: The grasshopper’s motion can be analyzed as projectile motion, where it follows a parabolic trajectory. The given information includes the launch angle (( \\theta_0 = 45^\\circ )) and the range (( R = 0.80 \\, \\text{m} )). Using the equation for the range of a projectile, we calculated the grasshopper’s initial velocity (( v_0 )). Since the angle of launch is ( 45^\\circ ), the range equation simplifies as ( \\sin 2\\theta_0 = 1 ).<\/p>\n\n\n\n With ( v_0 = 2.8 \\, \\text{m\/s} ), we determined the total time of flight (( T )) using the vertical motion equations. This time corresponds to the time the grasshopper is in the air during one hop. The horizontal component of the grasshopper’s velocity (( v_x )) remains constant throughout the hop because air resistance is neglected.<\/p>\n\n\n\n Finally, since the problem specifies that the time spent on the ground is negligible, the grasshopper’s average horizontal speed is the same as its horizontal velocity during the hop, ( v_x = 2.0 \\, \\text{m\/s} ). This result highlights the utility of decomposing projectile motion into horizontal and vertical components to analyze average speed and trajectory effectively.<\/p>\n","protected":false},"excerpt":{"rendered":" A grasshopper hops along a level road. On each hop, the grasshopper launches itself at angle \u03b80 = 45\u03bf and achieves a range R = 0.80 m. What is the average horizontal speed of the grasshopper as it hops along the road? Assume that the time spent on the ground between hops in negligible and […]<\/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-183907","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183907","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=183907"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183907\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183907"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183907"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183907"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Range of a projectile<\/h3>\n\n\n\n
[
R = \\frac{v_0^2 \\sin 2\\theta_0}{g}
]
where:<\/p>\n\n\n\n\n
[
0.80 = \\frac{v_0^2}{9.8}
]
[
v_0^2 = 0.80 \\times 9.8 = 7.84
]
[
v_0 = \\sqrt{7.84} \\approx 2.8 \\, \\text{m\/s}
]<\/p>\n\n\n\nStep 2: Time of flight<\/h3>\n\n\n\n
[
T = \\frac{2v_0 \\sin \\theta_0}{g}
]
Since ( \\theta_0 = 45^\\circ ) and ( \\sin 45^\\circ = \\frac{\\sqrt{2}}{2} ):
[
T = \\frac{2(2.8)(\\frac{\\sqrt{2}}{2})}{9.8} = \\frac{2.8 \\sqrt{2}}{9.8}
]
[
T \\approx 0.4 \\, \\text{seconds}
]<\/p>\n\n\n\nStep 3: Average horizontal speed<\/h3>\n\n\n\n
[
v_x = v_0 \\cos \\theta_0
]
[
v_x = 2.8 \\cdot \\cos 45^\\circ = 2.8 \\cdot \\frac{\\sqrt{2}}{2} \\approx 2.0 \\, \\text{m\/s}
]<\/p>\n\n\n\n
[
\\text{Average horizontal speed} = v_x \\approx 2.0 \\, \\text{m\/s}.
]<\/p>\n\n\n\nExplanation (300 words)<\/h3>\n\n\n\n