SOLUTION: 106. Pole vaulting. In 1981 Vladimir Poliakov (USSR) set a world record of 19 ft 3/4 in. for the pole vault (www.polevault.com). To reach that height, Poliakov obtained a speed

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Travel -> SOLUTION: 106. Pole vaulting. In 1981 Vladimir Poliakov (USSR) set a world record of 19 ft 3/4 in. for the pole vault (www.polevault.com). To reach that height, Poliakov obtained a speed      Log On

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 Click here to see ALL problems on Travel Word Problems Question 191944This question is from textbook Elementary and Intermediate Algebra : 106. Pole vaulting. In 1981 Vladimir Poliakov (USSR) set a world record of 19 ft 3/4 in. for the pole vault (www.polevault.com). To reach that height, Poliakov obtained a speed of approximately 36 feet per second on the runway. The formula h _ _16t2 _ 36t gives his height t seconds after leaving the ground. a) Use the formula to find the exact values of t for which his height was 18 feet. b) Use the accompanying graph to estimate the value of t for which he was at his maximum height. c) Approximately how long was he in the air? This question is from textbook Elementary and Intermediate Algebra Answer by Alan3354(30993)   (Show Source): You can put this solution on YOUR website!Pole vaulting. In 1981 Vladimir Poliakov (USSR) set a world record of 19 ft 3/4 in. for the pole vault (www.polevault.com). To reach that height, Poliakov obtained a speed of approximately 36 feet per second on the runway. The formula h _ _16t2 _ 36t gives his height t seconds after leaving the ground. h = -16t^2 + 36t a) Use the formula to find the exact values of t for which his height was 18 feet. 18 = -16t^2 + 36t 8t^2 - 18t + 9 = 0 t = 0.75 seconds (going up) t = 1.5 seconds (coming down) t found using quadratic equation. If you have questions, email me via the thank you note. b) Use the accompanying graph to estimate the value of t for which he was at his maximum height. I don't see a graph, but the easiest way to find max height is to get the average of the 2 values of t above. t = 1.125 seconds c) Approximately how long was he in the air? That would be 2 times the time at max height, or 2.25 seconds.