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 `10m//s^(2),` calculate
(i) time taken by the ball to reach the ground
(ii) velocity of the ball on reaching the ground
(iii) the height of the ball at half the time if takes to reach the ground.

Explanation

Given :
m = 3 kg, distance travelled by the ball (s) = 125 m, initial velocity of the ball (u) = 0 and acceleration (a) = g = `10m//s^(2).`
(i) Newton.s second equation of the motion gives,
`s=ut+(1)/(2)at^(2)`
`therefore125=0t+(1)/(2)xx10xxt^(2)`
`125=5t^(2)`
`t^(2)=(125)/(5)=25,t=5s`
The ball takes 5 seconds to reach the ground.
(ii) According to Newton.s first equation of motion final velocity,
v = u + at
`=0+10xx5`
= 50 m/s
The velocity of the ball on reaching the ground is 50 m/s
(iii) Half time = t = `(5)/(2)=2.5s`
Ball.s height at this time = s According to Newton.s second equation,
`s = ut+ (1)/(2)at^(2)`
`s=0+(1)/(2)xx10xx(2.5)^(2)=31.25m.`
Thus the height of the ball at half time = 125 - 31.25 = 93.75 m