Using_Quadratic_Equation

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Embry-Riddle Aeronautical University *

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111

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Physics

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Dec 6, 2023

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docx

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Uploaded by nataliemcginnes

Using a Quadratic Equation Problems A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s (t) = 112 + 96t - 16t 2 Complete the table and discuss the interpretation of each point. t s(t) Interpretation 0 112 At 0 sec the ball was at original height of 112ft. 0.5 156 At 0.5 sec (t) the ball will reach the height of 156ft. 1 192 At 1 sec the ball is at 192ft. 2 240 At 2 sec the ball is 240ft. -0.1225 100 A parabola graph, the ball will have two different y coordinates at 100ft. At -0.1225 seconds the ball will reach 100ft. It would take 0.1225 to reach 12ft if the ball was never thrown and released from its position. 6.1225 100 A Parabola graph, the ball will have two different y coordinates at 100ft. For the ball to reach 100ft it will take 6.1225 seconds after thrown upwards. 1.1292 200 The ball will reach the height of 200ft twice, once after being thrown and the other on its downward decent. After being released, it will take 1.1292 seconds to cross 200ft. 4.8708 200 The ball will reach the height of 200ft twice again, once after thrown and the other on its decent. After released, it will take 4.8708 seconds for the ball to reach its maximum height and fall back to 200ft. All rights are reserved. The material contained herein is the copyright property of Embry-Riddle Aeronautical University, Daytona Beach, Florida, 32114. No part of this material may be reproduced, stored in a retrieval system or transmitted in any form, electronic, mechanical, photocopying, recording or otherwise without the prior written consent of the University. College of Arts & Sciences | worldwide.erau.edu

Answer these questions. 1. After how many seconds does the ball strike the ground? s ( t ) = 112 + 96 ( 7 )− 16 ( 7 ) 2 s ( t ) = 112 + 672 − 16 × 49 s ( t ) = 112 + 672 − 784 s ( t ) = 0 Thus, at t=7 sec the ball will strike the ground. 2. After how many seconds will the ball pass the top of the building on its way down? 112 = 112 + 96 t − 16 t 2 96 t − 16 t 2 = 0 t ( 96 − 16 ) = 0 96 = 16 t t = 6 After t=6 sec the ball will pass the top of the building. 3. How long will it take the ball to reach the maximum height? s ( t ) = 112 + 96 ( 3 ) − 16 ( 3 ) 2 s ( t ) = 112 + 288 − 16 × 9 s ( t ) = 112 + 288 − 144 s ( t ) = 256 ft After t=3 sec the ball will reach the maximum height. Page 2 of 3 College of Arts & Sciences | worldwide.erau.edu All rights are reserved. The material contained herein is the copyright property of Embry-Riddle Aeronautical University, Daytona Beach, Florida, 32114. No part of this material may be reproduced, stored in a retrieval system or transmitted in any form, electronic, mechanical, photocopying, recording or otherwise without the prior written consent of the University .

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