A ball of mass 0 . 5 kg is dropped from the height of 10 m . The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is _____ m . [Use g = 10 m s - 2 ]

Step-by-step solution

Consider downward direction to be positive. Applying first equation of motion, v = u + a t ⇒ v = 0 + g t   ⇒ g = g t   ⇒ t = 1   s Distance covered in 1   s , s = u t + 1 2 a t 2 = 0 + 1 2 g × 1 2 = 10 2 = 5   m Hence, height of the point from the surface = 10 - 5   m = 5   m