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Unformatted text preview: of your results explaining them in the context of the original problem. Consider a ball thrown from a height of 10ft with initial velocity 2 m/s So it means v = 2 m/s , k = 100 ft, So the formula of height at any time t becomes S=16t 2 +2t + 100 Now consider t = 2 seconds S = 16(2) 2 + 2(2) + 100 S = 64 + 4 + 100 S = 40 ft So after 2 seconds the ball is 40ft above the ground Consider t = 1 seconds S = 16(1) + 2(1) + 100 S = 86 ft above the ground So from above we see that at t = 2 seconds the ball is at 40 ft height while at t =1 second it is 86 ft, since the ball is thrown from 100 ft so its obvious that at some time later it will be at a lower depth Now the height becomes 0 when 16t 2 + 2t+ 100 = 0 Which on solving gives t = 2.56 seconds So at t = 2.56 seconds the ball will reach the ground Here the is graph which shows the trajectory of the ball...
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This note was uploaded on 02/20/2011 for the course ALGEBRA 1 taught by Professor Taylor during the Spring '11 term at American InterContinental University.
 Spring '11
 Taylor
 Algebra

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