An object is dropped from a vertical height of 1.89 m above the balcony level. What is the object’s speed when it is 2.20 m below the balcony level if 10.0% energy is lost due to the air resistance? Does it matter when to apply 10% loss before V calculations or after? [8.49m/s] [yes it does, 0.9Energy result in √0.9Velocity]

The 10% must come off the energy, not the velocity

Explanation:

Change in height h = 1.89 + 2.20 = 4.09 m

Potential energy = mgh = m(9.81)(4.09) = 40.1m J

Friction losses 40.1m(0.10) = 4.01m J

Kinetic energy at end 40.1m - 4.01m = 36.1m J

KE = ½mv²

v = √(2KE/m)

v = √(2(36.1m)/m = √72.2

v = 8.4983...

v = 8.50 m/s

a.   mass number from atomic number