From the top of a 64 metres high tower, a stone is thrown upwards vertically with the velocity of 48 m/s. The greatest height (in metres) attained by the stone, assuming the value of the gravitational acceleration $$g = 32$$ m/s$$^2$$, is:

We are given that the stone is projected vertically upward from the top of a tower of height $$64\ \text{m}$$ with an initial velocity $$u = 48\ \text{m s}^{-1}$$. The acceleration due to gravity acts downward with magnitude $$g = 32\ \text{m s}^{-2}$$. We wish to find the greatest height that the stone reaches above the ground.

First we determine how high the stone rises above the point of projection (the tower-top). At the highest point, the vertical velocity becomes zero. Using the first equation of motion, which in general form is written as

$$v = u - g t,$$

we set the final velocity $$v = 0$$ (because at the greatest height the stone momentarily comes to rest). Substituting $$u = 48\ \text{m s}^{-1}$$ and $$g = 32\ \text{m s}^{-2}$$, we get

$$0 = 48 - 32 t.$$

Solving for the time $$t$$ it takes to reach the maximum point:

$$32 t = 48 \quad\Longrightarrow\quad t = \frac{48}{32} = \frac{3}{2} = 1.5\ \text{s}.$$

Now we calculate the vertical displacement $$s$$ in this time interval. For vertical motion with constant acceleration, the displacement is obtained from the second equation of motion, stated as

$$s = u t - \frac{1}{2} g t^2.$$

Substituting the known values $$u = 48\ \text{m s}^{-1},\ t = 1.5\ \text{s},\ g = 32\ \text{m s}^{-2},$$ we have

$$\begin{aligned} s &= 48 \times 1.5 \;-\; \frac12 \times 32 \times (1.5)^2 \\ &= 72 \;-\; 16 \times 2.25 \\ &= 72 \;-\; 36 \\ &= 36\ \text{m}. \end{aligned}$$

Thus the stone rises $$36\ \text{m}$$ above the top of the tower.

The tower itself is $$64\ \text{m}$$ high, so the greatest height above ground level reached by the stone is

$$\text{Maximum height} = 64\ \text{m} + 36\ \text{m} = 100\ \text{m}.$$

Hence, the correct answer is Option D.