A ball is dropped from a height of 18 m and its velocity when it reaches a height of 3 m has to be determined.

Assume the total energy of the ball is conserved as it travels down. At a height of 18 m, it is at rest. The potential energy of the ball is m*18*9.8 and the kinetic energy is 0. At a height of 3 m, the potential energy of the ball is m*9.8*3 and the kinetic energy is (1/2)*m*v^2 where v is the speed of the ball.

18*9.8*m + 0 = 3*9.8*m + (1/2)*m*v^2

=> 18*9.8 = 3*9.8 + (1/2)*v^2

=> v^2 = 294

=> v = 17.14 m/s

The speed of the ball at a height of 3 m is 17.14 m/s

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