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
he height of the ball at half the time it takes to reach 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)`.
Half time `=t=(5)/(2) =2.5 s`
Ball’s height at this time = s
According to Newton’s second equation
`s=ut +(1)/(2) at^(2)`
`s=0+(1)/(2) 10xx(2.5)^(2) =31.25 m`.
Thus the height of the ball at half time
= 125-31.25 = 93.75 m