Question

What is the mass of a baseball if the kinetic energy is 122 J when the baseball is thrown at 103 mph? (in grams)

A batter hits a pop fly, and the baseball reaches an altitude of 230 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? (in joules)

Answer 1

Kinetic energy of the base ball K.E.= 122J= 122X 10⁷ ergs . ( 1 J = 10⁷ ergs )

Velocity of the base ball, V=103 mph = 103 X 44.704 cm/s =4604.512cm/s. (1 mph = 44.704 cm/s)

We know that. K.E.= 1/2 mV²

122 X 10⁷ = 1/2 m (4604.513)²

m = (122 X 10⁷ X 2)/(4604.512)² = 115.086 gm.

Mass of the base ball, m= 115.086 gm= 115.086 X 10^-3 kg.

The acceleration due to gravity, g = 9.8 m/sec²

Initial position of the base ball, h₁ = 3 ft. = 3 X 0.305 m ( 1 ft = 0.305 m )

Initial P.E of the base ball, P.E₁ = m g h₁ = 115.086 X 10^-3 X 9.8 X 0.915 = 1.032 J

Final position of the base ball, h₂= 230 ft = 230 X 0.305 = 70.15 m

Final P.E. of the base ball, P.E₂ = m g h₂ = 115.086 X 10^-3 X 9.8 X 70.15 = 79.12 J

The increase in potential energy of the base ball = P.E₁ - P.E₂ = 79.12- 1.032 = 78.08 J.