# Application of conservation of mechanical energy for projectile motion – problems and solutions

1. A kicked football leaves the ground at an angle θ = 30o with the initial velocity of 10 m/s. Ball’s mass = 0.1 kg. Acceleration due to gravity is 10 m/s2. Determine (a) The gravitational potential energy

at the highest point (b) The highest point or the maximum height

Known :

Mass (m) = 0.1 kg

The initial velocity (vo) = 10 m/s

Angle = 30o

Acceleration due to gravity (g) = 10 m/s2

Solution :

(a) The gravitational potential energy Calculate the horizontal component (vox) and the vertical component (voy) of initial velocity.  vox = vo cos θ = (10)(cos 30o) = (10)(0.5√3) = 5√3 m/s

voy = vo sin θ = (10)(sin 30o) = (10)(0.5) = 5 m/s

The initial mechanical energy

The initial mechanical energy (MEo) = kinetic energy (KE)

MEo = KE = ½ m vo2 = ½ (0.1)(10)2 = ½ (0.1)(100) = ½ (10) = 5 Joule

The final mechanical energy

Kinetic energy at the highest point :

KE = ½ m vox2 = ½ (0.1)(5√3)2 = ½ (0.1)((25)(3)) = ½ (0.1)(75) = 3.75 Joule

Principle of conservation of mechanical energy

The initial mechanical energy (MEo) = the final mechanical energy (MEt)

KE = PE + KE

5 = EP + 3.75

PE = 5 – 3.75 = 1.25 Joule

The gravitational potential energy at the highest point is 1.25 Joule.

(b) The highest point or the maximum height

PE = m g h

1.25 = (0.1)(10) h

1.25 = h

The maximum height is 1.25 meters.

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2. A 0.1-kg ball projected horizontally with initial velocity vo = 10 m/s from a building 10 meter high. Acceleration due to gravity is 10 m/s2. Determine ball’s kinetic energy when it hits the ground.

Known :

Mass (m) = 0.1 kg

Initial velocity (vo) = 10 m/s

Acceleration due to gravity (g) = 10 m/s2

The change in height (h) = 10 – 2 = 8 m