A basebal (radius =.036 m, mas =.145kg) is droped from rest at the top of the Empire State Building (height = 1250ft). Calculate (a) the initial potential energy of the basebal, (b) its final kinetic energy.

a) the initial potential energy = 541.95J

b) the final kinetic energy = 87.991 J

Explanation:

Step 1: Data given

The ball has a mass of 0.145 kg and is at a height of 381m

Step 2: Calculate potential energy

The potential energy = m * g * h

with m = the mass of the ball = 0.145 kg

with g = Gravitational acceleration = 9.81 m/s²

with h = the height of the building = 381m

The potential energy = 0.145 Kg * 9.81 m/s² * 381m = 541.95 J

FD = 1/2ρCDAv²

⇒ with FD = the drag force = the force component in the direction of the flow velocity

⇒ with ρ = density of the fluid (air in our case: ρ≈1.1839 Kg/m3 at 1 atm and 25 °C)

⇒ with v = velocity of the ball

⇒ with A = reference area, which in our case is just the cross sectional area of the ball: A=πr2

⇒ with CD is the drag coefficient - a dimensionless coefficient, that in the case of a sphere, CD=0.47

Following Newton's second law:

ΣFy = may = - mg + Dv²

Here is D=1/2ρCDA (for convenience) = 0.001172

The terminal speed we can define as the speed of the ball where ay = 0

Therefore: - mg + Dvt² = 0

⇒vt = √ (mg/D)

vt = √ (0.145 * 9.8 / 0.001172) ≈ 34.837

Uinitial=mgh≈541.951 J (see first question)

Kfinal=1/2*mvt² = (m²g) / 2D ≈ 87.991 J

The final kinetics energy is 87.991 J