From 18 m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is _____ m. (Take g = 10 m/s ^2 and neglect the air resistance)

Step-by-step solution

Let the ball fall a distance s from the initial height. Initial velocity u = 0 Acceleration a = g = 10 m/s ^2 Given that the magnitude of velocity is equal to the magnitude of acceleration due to gravity, we have v = 10 m/s. Using the equation of motion v^2 = u^2 + 2gs : (10)^2 = 0^2 + 2 10 s 100 = 20s s = 5 m The distance fallen by the ball is 5 m. The height above the ground is h = H - s h = 18 - 5 = 13 m Answer: 13