If a body at rest is dropped from a certain height and takes 4 seconds to reach the ground, what is the approximate height from which it was dropped? (Take g = 10 m/s²)

The correct option is 80 meters.

To find the height from which the body was dropped, we can use the second equation of motion under gravity:
$$s=ut+\frac{1}{2}g{t}^2$$

Let's identify the given values from the problem:
1. Since the body starts from rest, its initial velocity (u) is $0\text{m/s}$.
2. The time taken (t) to reach the ground is $4\text{seconds}$.
3. The acceleration due to gravity (g) is given as $10{\text{m/s}}^2$.
4. The displacement (s) represents the height (h) from which the body was dropped.

Now, substitute these values into the equation:
$$h=(0)(4)+\frac{1}{2}(10)(4^2)$$

Simplify the terms step-by-step:
$$h=0+5\times 16$$
$$h=80\text{meters}$$

Therefore, the approximate height from which the body was dropped is 80 meters.