A ball is dropped from the top of a building of height 80 m. Considering the acceleration due to gravity of the Earth to be 10 m/s², the velocity of the ball right before it touches the ground is ____ m/s.

The correct answer is: 40

Given that,
The initial velocity, u = 0
The acceleration, a = 10 m/s²
The displacement, s = 80 m
On using the third equation of motion we get,
v² = u² + 2as
Or, (v)² = 0 + 2 x 10 x 80
Or, (v)² = 1600
Or, v = √(1600)
Or, v = 40 m/s
Therefore, the velocity of the ball before hitting the ground is 40 m/s.