A stone thrown upward with a speed u from the top of the tower reaches the ground with a velocity 3u. The height of the tower is

The correct option is 4u² / g.

Let us analyze the motion of the stone step-by-step using the equations of motion under gravity.
Let:
- $u$ be the initial velocity of the stone, thrown vertically upward.
- $v$ be the final velocity of the stone when it reaches the ground.
- $h$ be the height of the tower.
- $g$ be the acceleration due to gravity, acting vertically downward.

Let us choose the vertically downward direction as positive. Based on this sign convention:
- The initial velocity of the stone is upward, so its sign is negative: $v_i=-u$.
- The final velocity of the stone on reaching the ground is downward, so its sign is positive: $v_f=3u$.
- The displacement of the stone is downward, from the top of the tower to the ground, so its sign is positive: $s=+h$.
- The acceleration due to gravity is downward, so its sign is positive: $a=+g$.

We can use the third equation of motion, which relates initial velocity, final velocity, acceleration, and displacement:
${v_f}^2-{v_i}^2=2as$

Substituting the values with their respective signs into the equation:
${\left(3u\right)}^2-{\left(-u\right)}^2=2gh$

Simplifying the squared terms:
$9u^2-u^2=2gh$

Combine the terms on the left side:
$8u^2=2gh$

Solving for the height of the tower, $h$:
$h=\frac{8u^2}{2g}$
$h=\frac{4u^2}{g}$

Thus, the height of the tower is $\frac{4u^2}{g}$.