A ball is thrown directly downward, with an initial speed of 6.10 m/s, from a height of 31.0 m. After what interval does the ball strike the ground?

Answer 1

d = Vi*t + 0.5gt^2 = 31m,
6.1t + 0.5*9.8t^2 = 31,
4.9t^2 + 6.1t - 31 = 0,
Solve for t using Quad. Formula and get:

t = 1.97s.