A ball of mass 10kg is dropped from a height of 50m, if the work done against drag is 500J what is the speed of the ball immediately before impact with the floor? (g=10N/Kg)
To solve this problem we will use the principle of conservation of energy. The gravitational potential energy of the ball at the maximum height must be equal to the kinetics energy at the bottom plus the work done against drag. The GPE is given by the equation GPE=mgh. Therefore in this case it is 101050=5000J.If we subtract the work against drag from this we will be left with the balls kinetic energy at the bottom of the fall. 5000-500=4500.The kinetic energy of an object is given by 1/2mv^2. We must then rearrange this for v and after putting in the values the speed is 30m/s. It is important to always include units in the final answer.
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