A ball is kicked directly upward. After
the ball reaches its highest altitude, it falls back down to its initial
position because of gravity. The altitude function of the
ball is s = 30t  5t^{2}. How high can the ball go up? The for displacement and velocity units are m and m/s, respectively.
Since the ball starts and ends at the same vertical location, its displacement
function is continuous and differentiable. This satisfies the prerequisite of Rolle's Theorem, and thus a point exists so that the derivative with
respect to time, t, is 0.
ds/dt = d(30t 5t^{2})dt
=
d(30t)/dt  d(5t^{2})/dt
= 30  10t
As mentioned this must equal zero,
ds/dt = 30  10t = 0
Since the derivative of displacement with respect to time is velocity,
the above equation can be rewritten as
ds/dt = v =
30  10t = 0
Therefore, t = 3
Substituting t = 3 into the displacement function
s = 30t  5t^{2}
= 30(3)  5(3)^{2}
= 45 m
Thus, the maximum altitude of the ball will be 45 m.
