Suppose you launch a water balloon...?

Question

straight into the air from the ground and determine that it reached a maximum height of 21.16 feet. You further determine that it hit the ground 2.3 seconds after launch. When was the balloon 10 feet from the ground?

Answer 1

Vo = √[(2)(32.2)(21.16)]

- 16.1t^2 + t√[(2)(32.2)(21.16)] - 10 = 0

t = {√[(2)(32.2)(21.16) ± √[(2)(32.2)(21.16) - (2)(32.2)(10)]}/32.2

t = {√[(2)(32.2)(21.16) ± √[(2)(32.2)(11.16)]}/32.2

t = [√(1/16.1)](√21.16 ± √11.16)

t ≈ 0.31386 s, 1.9790 s

t ≈ 0.314 s, 1.98 s