A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s⁻², with what velocity will it strike the ground?

Height, s = 20 m
Acceleration, a = 10 m/s²
By the 3rd equation of motion:
v² = u² + 2as
v² = 0 + 2 × 10 × 20 = 400 ⇒ v = 20 m/s
So, we conclude the final velocity, v = 20 m/s
Again, by the 1st equation of motion, v = u + at ⇒ t = (v - u) / a ⇒ t = 20 / 10 = 2 s
So, it will strike the ground after 2 s