1. The problem statement, all variables and given/known data An Olympic long jumper leaves the ground at an angle of 23o and travels through the air for a horizontal distance of 8.7m before landing. What is the takeoff speed of the jumper? 2. Relevant equations 8.7m = cos(23o)Vot y = sin(23o)t + (1/2)(-9.8m/s2)t2 0m2/s2 = sin2(23o)Vo2 + 2(-9.8m/s2)y 3. The attempt at a solution Fifteen minutes and many permutations later, I get t = .835, .782 with Vo = 11.32m/s, 12.09m/s The answer in the book is 11m/s. Can you explain the steps you would take to solve this? It literally took me 15 minutes of mathematical manipulation to isolate t. The final step to find t for me was: .697 = t2(3.6934 - 4.9t2) Thanks guys!