A body released from a height h takes time t to reach earth’s surface. The time taken by the same body released from the same height to reach the moon’s surface is

The correct option is √6 t.

To find the time taken by a body to reach the moon's surface when released from a height
$h$
, we can use the equations of motion under gravity.

For a body falling freely from rest (where initial velocity
$u=0$
) from a height
$h$
, the distance traveled
$s=h$
in time
$t$
is given by the second equation of motion:

$s=ut+\frac{1}{2}g{t}^2$

Since the body is released from rest (
$u=0$
):

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

Solving for time
$t$
, we get:

$t=\sqrt{\frac{2h}{g}}$

1. On Earth's Surface:
Let the acceleration due to gravity on Earth be
$g_e$
and the time taken be
$t$
.
$t=\sqrt{\frac{2h}{g_e}}$

2. On the Moon's Surface:
Let the acceleration due to gravity on the Moon be
$g_m$
and the time taken be
$t_m$
. Since the body is released from the same height
$h$
:
$t_m=\sqrt{\frac{2h}{g_m}}$

We know that the acceleration due to gravity on the Moon is approximately one-sixth of that on Earth:
$g_m=\frac{g_e}{6}$

Substituting this relationship into the expression for
$t_m$
:

$t_m=\sqrt{\frac{2h}{\left(\frac{g_e}{6}\right)}}$

Simplifying the fraction:

$t_m=\sqrt{6\times \frac{2h}{g_e}}$

$t_m=\sqrt{6}\times \sqrt{\frac{2h}{g_e}}$

Since the term under the second square root is the time taken on Earth (
$t$
), we can substitute it back:

$t_m=\sqrt{6}t$

Therefore, the time taken to reach the Moon's surface is √6 t.