When a ball is thrown upwards it rises to a height of 18 m before falling down. The speed at which the ball was thrown up has to be determined.

The total energy of a closed system is constant. Here, the ball can be considered to form a closed system. The total energy of the ball when it is at ground level and when it is at the highest point it rises to is the sum of its potential and kinetic energy. The gravitational potential energy of the ball at height h is m*g*h where the mass of the ball is m and g is the gravitational acceleration. The kinetic energy of a ball moving at velocity v is `(1/2)*m*v^2` . Let the speed at which the ball is thrown upwards be v. At the highest point the speed of the ball is 0. Equating the total energy of the ball gives:

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

=> `v^2 = 2*9.8*18`

=> v `~~` 18.78 m/s

The ball is thrown upwards with a speed approximately equal to 18.78 m/s

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