A person drops a brick from the top of a building. The height of the building is 400 m and the mass of the brick is 2.00 kg. What will be the speed of the brick right before it touches the ground? Use g=10.0 m/s^2.

This question involves the conservation of energy. There are two energy in this case, potential energy and kinetic energy. Let's divid the energy into three status. 1. Before dropping, all potential energy 2.dropping, potential energy transformed to kinetic energy3. before hitting the ground, all Kinetic energy.Recall the formula for both energy, which are U=mgh, and K=1/2mv^2Since the energy is conserved in this case ( b/c otherwise it will say in the problem), the amount of energy at the beginning should equal to the energy at the end. Therefore we have, mgh=1/2mv^2plug the number in and solve for velocity.2x400x10=1/2 x 2 x v^2v^2=8000v=[tex] \sqrt{8000} [/tex]v=40[tex] \sqrt{5} [/tex]