A ball at rest is dropped from the top of a 20 m high building. Calculate the speed with which it hits the ground. (Take g = 10 m/s²)

The correct option is 20 m/s.

To find the speed with which the ball hits the ground, we can use the equations of motion under uniform acceleration. Let us list the given information from the problem:

1. Initial velocity (since the ball starts from rest), u = 0 m/s
2. Height of the building (distance traveled), s = 20 m
3. Acceleration due to gravity, g = 10 m/s²

We need to find the final velocity (speed), v, when the ball hits the ground. We can use the third equation of motion:

$$v^2=u^2+2gs$$

Substituting the given values into the equation:

$$v^2={\left(0\right)}^2+2\times 10\times 20$$

Simplifying the right side of the equation:

$$v^2=0+400$$

$$v^2=400$$

Taking the square root of both sides to find v:

$$v=\sqrt{400}$$

$$v=20\text{ m/s}$$

Therefore, the speed with which the ball hits the ground is 20 m/s.