Answer: 4.52 seconds
Step-by-step explanation:
We know that acceleration is the rate of change of the velocity, and velocity is the rate of change of the position.
If we define h(t) as the position, then we have that
h'(t) = velocity
h''(t) = acceleration
And we know that the acceleration is equal to -9.8m/s^2
then we have:
h''(t) = -9.8m/s^2 and the initial condition h(0) = 100m
For solving this, we integrate two times with respect to the time:
h'(t) = (-9.8m/s^2)*t + h'(0)
where h'(0) is the initial velocity, that in this case is 0m/s because the ball is droped.
We integrate again:
h(t) = ((-9.8m/s^2)/2)*t^2 + h(0) = -4.9m/s^2*t^2 + 100m
we need to find the time t in which h(t) = 0
0 = -4.9m/s^2*t^2 + 100m
4.9m/s^2*t^2 = 100m
t = (√100/4.9)s = 4.52 seconds