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

We are using the formula right here. $v^2-u^2=2as$ Keeping in mind that at the highest speed,$v=0.$ then ascertain the tower's height.

Let u be the stone's starting velocity, v its ultimate velocity, an its acceleration caused by gravity, s its displacement, and h its highest point.

Think of a stone falling at a certain speed of $48m{s}^{-1}$ from a building that is 64 meters high. .

We need to find the greatest height attained by the stone, assuming $g=32m{s}^{-2}$.

As the stone is falling down $g=-32m{s}^{-2}$.

WKT, according to kinematic equations:

$v^2=u^2+2as$

Where, v is final velocity and u is initial velocity.

Here, ⁣$v=0 u=48m{s}^{-1} a=g=-32m{s}^{-2}$

Substituting the values we get,

$$\begin{gathered} 0={\left(48\right)}^2+2\times \left(-32\right)\times s \\ 0=2304-64s \\ 64s=2304 \\ s=\frac{2304}{64}=36m \end{gathered}$$

Thus, the greatest height would be, 64+36=100m