A brick dropped from 78m high scaffold. what is the velocity as the brick hits the pavement? show work.

39.09987212 m/s There are a couple of ways to solve this problem. We can use the equation for the distance an object moves under a constant acceleration of 9.8 m/s^2, or we can calculate the kinetic energy an object will have after falling 78 meters and from there calculate the velocity. Let's first use the constant acceleration formula: d = 0.5 AT^2 78 m = 0.5 * 9.8 m/s^2 * T^2 78 m = 4.9 m/s^2 * T^2 15.91836735 s^2 = T^2 3.98978287 = T So we now know it takes 3.98978287 second for the brick to hit the ground. And at an acceleration of 9.8 m/s^2 we get: 3.98978287 s * 9.8 m/s^2 = 39.09987212 m/s Now let's try using the energy method. I will assume that the mass of the brick is 1 kg, but any non-zero value will do. So 78 m * 9.8 m/s^2 * 1 kg = 764.4 kg*m^2/s^2 And kinetic energy is expressed as KE = 0.5 mv^2 So 764.4 kg*m^2/s^2 = 0.5 1 kg * v^2 764.4 kg*m^2/s^2 = 0.5 kg * v^2 1528.8 m^2/s^2 = v^2 39.09987212 m/s = v So the determine how long it takes and multiply by the acceleration method gives 39.09987212 m/s. And the use the energy to determine the velocity gives 39.09987212 m/s. So it doesn't matter which of the two methods you use. But, personally, I think the energy method is simpler, although the time method is more direct.