Question

For the situation depicted in the Fig, calculate the mechanical energy of the ball just before it hits the ground and show that even at this position, it is mgh.

Answer 1

At point A, the ball is dropped from a height h with an initial velocity of u = 0. Its height drops to zero at point C, right before it touches the earth.

Let v be the ball's speed right before it reaches the ground.

By applying the motion equation,

v² = u² + 2gh

v² = 0 + 2gh = 2gh

At this position (just before hitting the ground):

Kinetic energy = `1/2 "mv"^2 = 1/2 xx m xx 2gh = mgh`

Potential energy = mg × 0 = 0 (since height = 0)

Mechanical energy = Kinetic energy + Potential energy

Mechanical energy = mgh + 0 = mgh

As a result, the ball's mechanical energy right before it strikes the ground is mgh, which is equal to its original mechanical energy at point A. This shows the conservation of mechanical energy.