An iron ball of mass 3 kg is released from a height of 125 m and falls freely to the ground. Assuming that the value of g is `10 m//s^(2)`, calculate
velocity of the ball on reaching the ground

Explanation

m = 3 kg, distance travelled by the ball s = 125 m, initial velocity of the ball = u = o and acceleration `a = g = 10 m//s^(2)`.
According to Newton’s first equation of motion final velocity = v = u + a t
`=0+10xx5`
`=50 m//s`
The velocity of the ball on reaching the ground is 50 m/s