When a ball of mass m is dropped from rest from a height h, its kinetic energy just before landing is k. now, suppose a second ball of mass 4m is dropped from rest from a height h/4?

The energy will be the same. To give you some concrete numbers to support this, let's drop something and see where the math takes us. Under constant acceleration, the distance an object moves is given by d = 0.5 A T^2 Now if you double T, you'll see that the distance quadruples. Since d = 0.5 A (2T)^2 d = 0.5 A 4T^2 d = 2AT^2 and 2AT^2/(0.5 AT^2) = 2/0.5 = 4 So dropping from 4 times the height doubles the time it takes to fall. And since the time is doubled, the velocity the object has when it hits is also doubled. And the formula for kinetic energy is E = 0.5MV^2 So let's plug in a few numbers. First object. Mass = M, velocity = 2V Second object. Mass = 4M, velocity = V Kinetic energy first object is: KE = 0.5M(2V)^2 = 0.5M4V^2 = 2MV^2 Kinetic energy second object is: KE = 0.5 4MV^2 = 2MV^2 And they're the same.

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