A body falls from rest in the gravitational field of the earth. The distance travelled in the fifth second of its motion is (g= 10m/s²)

The correct option is 45m.

To find the distance travelled by a body in a specific second of its motion, we can use the formula for the distance covered in the n-th second of uniformly accelerated motion:

$S_n=u+\frac{a}{2}(2n-1)$

Let's identify the given values from the problem statement:
1. Since the body falls from rest, the initial velocity is:
$u=0\text{m/s}$
2. The acceleration is due to gravity, acting downwards:
$a=g=10{\text{m/s}}^2$
3. We need to find the distance travelled in the fifth second of motion, so:
$n=5$

Now, substitute these values into the formula:

$S_5=0+\frac{10}{2}\left(2\right(5)-1)$

Simplify the expression inside the parentheses first:

$S_5=5(10-1)$

$S_5=5\left(9\right)$

$S_5=45\text{m}$

Therefore, the distance travelled by the body in the fifth second of its motion is 45 meters.