Question

A batter hits a pop fly, and the baseball (with a mass of 148 g) reaches an altitude of 265 ft. If we assume that the ball was 3 ft. above home plate when hit by the batter, what is the ball's increase in potential energy?

Answer 1

The increase in potential energy of the ball is 115.82 J

Step-by-step explanation:

Conceptual analysis

Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:

U = m × g × h

U: Potential Energy in Joules (J)

m: mass in kg

g: acceleration due to gravity in m/s²

h: height in m

Equivalences

1 kg = 1000 g

1 ft = 0.3048 m

1 N = 1 (kg×m)/s²

1 J = N × m

Known data

`h_2 = 265ft * (0.3048m)/(ft) = 80.77m`

`h_1 = 3ft * (0.3048m)/(ft) = 0.914m`

`m = 148g*(1kg)/(1000g) = 0.148kg`

`g = 9.8 (m)/(s^2)`

Problem development

ΔU: Potential energy change

ΔU = U₂ - U₁

U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁

U₂ - U₁ = mₓg(h₂ - h₁)

`U_2 - U_1 = 0.148kg * 9.8 (m)/(s^2)*(80.77m - 0.914m) = 115.82 N * m = 115.82J`

The increase in potential energy of the ball is 115.82 J