Originally Posted by

**ecMathGeek** We are given the following initial conditions:

y0 = 5 m

v0 = 100 m/s

y1 = 0 ... since we want it to hit the ground, we need its 'final' height to equal 0.

We'll use the distance formula:

y = y0 + v0*t + 1/2*g*t^2 ... where y0 is the initial height of the ball, v0 is the initial velocity, g is the acelleration of gravity = -9.8 m/s^2:

0 = 5 + 100t - 4.9t^2

4.9t^2 - 100t - 5 = 0

Now, use the quadratic formula to solve for t:

t = (100 + sqrt(10098))/9.8 = 20.46 s

NOTE: I did the positive version of the quadratic formula since the 'minus' version would give a negative number for t (which we don't need).