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 - 5t2. 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 -5t2)dt
d(30t)/dt - d(5t2)/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 - 5t2
= 30(3) - 5(3)2
= 45 m
Thus, the maximum altitude of the ball will be 45 m.