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

1. A kicked football leaves the ground at an angle θ = 30^{o }with the initial velocity of 10 m/s. Ball’s mass = 0.1 kg. Acceleration due to gravity is 10 m/s^{2}. 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 (v_{o}) = 10 m/s

Angle = 30^{o}

Acceleration due to gravity (g) = 10 m/s^{2}

__Solution :__

**(a) The gravitational potential energy**

Calculate the horizontal component (v_{ox}) and the vertical component (v_{oy}) of initial velocity.

v_{ox }= v_{o} cos θ = (10)(cos 30^{o}) = (10)(0.5√3) = 5√3 m/s

v_{oy }= v_{o} sin θ = (10)(sin 30^{o}) = (10)(0.5) = 5 m/s

**The initial mechanical energy**

The initial mechanical energy (ME_{o}) = kinetic energy (KE)

ME_{o }= KE = ½ m v_{o}^{2} = 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 v_{ox}^{2} = 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 (ME_{o}) = the final mechanical energy (ME_{t})

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.

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

__Known :__

Mass (m) = 0.1 kg

Initial velocity (v_{o}) = 10 m/s

Acceleration due to gravity (g) = 10 m/s^{2}

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

__Wanted:__ kinetic energy at 2 meters above the ground

__Solution :__

The gravitational potential energy (PE) = m g h = (0.1)(10)(10) = 10 Joule

The initial kinetic energy (KE)= ½ m v_{o}^{2 }= ½ (0.1)(10)^{2} = ½ (0.1)(100) = ½ (10) = 5 Joule

The final kinetic energy = the initial gravitational potential energy + the initial kinetic energy = 10 + 5 = 15 Joule

### Ebook PDF Application of conservation of mechanical energy for projectile motion

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