A particle is dropped under gravity from rest from a height h (g= 9.8 m/s² ) and it travels a distance 9h/25 in the last second, the height h is

The correct option is 122.5 m.

Let the total height from which the particle is dropped be $h$ and the total time taken to reach the ground be $t$ seconds.
Since the particle is dropped from rest, its initial velocity $u=0$.
Using the equation of motion for free fall, the total height $h$ is given by:

$h=\frac{1}{2}g{t}^2$

where $g=9.8{\text{m/s}}^2$ is the acceleration due to gravity.

The distance traveled in the last second (the $t$-th second) is given by the formula:
$S_t=u+\frac{1}{2}g\left(2t-1\right)$

Since $u=0$ and the distance traveled in the last second is given as $\frac{9h}{25}$, we can write:

$\frac{9h}{25}=\frac{1}{2}g\left(2t-1\right)$

Substituting the expression for $h$ from the first equation into this equation, we get:

$\frac{9}{25}\left(\frac{1}{2}g{t}^2\right)=\frac{1}{2}g\left(2t-1\right)$

Canceling $\frac{1}{2}g$ from both sides, we simplify the relation to:

$\frac{9}{25}{t}^2=2t-1$

Multiplying both sides by 25 to clear the fraction:
$9t^2=25\left(2t-1\right)$
$9t^2=50t-25$
Rearranging this into a standard quadratic equation format:
$9t^2-50t+25=0$

Solving the quadratic equation by factoring:
$9t^2-45t-5t+25=0$
$9t\left(t-5\right)-5\left(t-5\right)=0$
$\left(9t-5\right)\left(t-5\right)=0$

This gives two possible solutions for time:
$t=5\text{s}$ or $t=\frac{5}{9}\text{s}$

Since the motion includes a "last second", the total time $t$ must be greater than 1 second. Thus, we select:
$t=5\text{s}$

Now, we calculate the height $h$ by substituting $t=5\text{s}$ and $g=9.8{\text{m/s}}^2$ into our height formula:

$h=\frac{1}{2}\times 9.8\times 5^2$

$h=4.9\times 25$

$h=122.5\text{m}$