fbpx

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

Application of conservation of mechanical energy for projectile motion – problems and solutions 1

Calculate the horizontal component (vox) and the vertical component (voy) of initial velocity.

Application of conservation of mechanical energy for projectile motion – problems and solutions 2Application of conservation of mechanical energy for projectile motion – problems and solutions 2vox = 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.

See also  Electric charge stored in capacitor – problems and solutions

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

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 vo2 = 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

[wpdm_package id=’1173′]

  1. Work done by force problems and solutions
  2. Work-kinetic energy problems and solutions
  3. Work-mechanical energy principle problems and solutions
  4. Gravitational potential energy problems and solutions
  5. The potential energy of elastic spring problems and solutions
  6. Power problems and solutions
  7. Application of conservation of mechanical energy for free fall motion
  8. Application of conservation of mechanical energy for up and down motion in free fall motion
  9. Application of conservation of mechanical energy for motion on a curved surface
  10. Application of conservation of mechanical energy for motion on an inclined plane
  11. Application of conservation of mechanical energy for projectile motion

Print Friendly, PDF & Email

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Discover more from Physics

Subscribe now to keep reading and get access to the full archive.

Continue reading