Energy Conservation (Physics)

Question

A Tennis ball of 120 Kg is dropped from 5m height. Ignore the air resistance, What is the speed of the ball when it hits the ground.

Answer 1

We will analyze the situation at two instants:

1- The instant the ball is released from the roof

2- The instant the ball hits the ground

In instant 1, the Total Mechanical Energy (TME) of the system (ball) is in the form of Gravitational Potential Energy (PE_grav) as it is at rest (no kinetic energy) and it is elevated above the surface of the Earth.

TME₁= PE_{grav................................{eq. 1}}

Gravitational potential energy can be described by the following equation:

PE_grav= m*g*h [J].............................{eq. 2}

Where:

m= Mass of the object [kg]

g= Acceleration due to gravity [m/s²]

h= Height above the reference plane (surface of the Earth) [m]

In instant 2, the TME of the system is in the form of Kinetic Energy (KE) because it is in motion and it has no gravitational potential energy (because it is on the surface of the Earth)

TME₂= KE...........................{eq. 3}

Kinetic Energy can be described by the following equation:

KE= 0.5* m * v².................................{eq. 4}

With:

m= Mass of the object [kg]

v= Velocity of the object [m/s]

Due to the Conservation of Energy, we know that the Total Mechanical Energy at instant 1 and instant 2 is conserved, in other words, they are equal in magnitude.

TME₁ = TME₂.................................{eq. 5}

Substituting equations 1 and 3 into equation 5:

PE_grav = KE..........................{eq. 6}

Substituting equations 2 and 4 into equation 6:

m*g*h = 0.5*m*v²

Solving for v:

v = (2*g*h)^1/2

Substituting g=10 m/s², h= 5 m:

v = (2*10*5)^1/2

v= 10 m/s