A block is initially at rest at the top of a 8.0 m high hill. The block slides down to the bottom of the hill and the back up a second hill with a height of 5.4 m. What is the
speed of the block at the top of the second hill. Assume no frictional forces.

The speed of the block at the top of the second hill = 7.14 m/s

Explanation:

According to the principle of conservation of energy, the potential energy lost by the block due to decrease in elevation is equal to the kinetic energy gained by the block due to motion

The initial potential energy of the block is found from the equation of potential energy as follows;

Where the condition of no friction  is applied

PE = Mass, m× Acceleration due to gravity, g × Height, h

Where:

m = The mass of the block

g = 9.81 m/s²

h is given as 8.0 m

The kinetic energy, KE, gained by the sliding block = 1/2 × mass of the block, m × (Velocity of the block, v)²

That is KE = 1/2×m×v²

At the bottom of the hill the potential energy is completely converted into potential energy which gives;

m × 9.81 × 8 = 1/2×m×v²

KE = 1/2×m×v² = m×78.48

9.81 × 8 = 1/2×v² (By cancelling the like term on both sides of the above equation)

v² = 2 × 9.81 × 8 = 156.96

v = √156.96 = 12.53 m/s

At the height h = 5.4 m. of the second hill, we have;

Potential energy gained = Kinetic energy lost

Kinetic energy remaining = Initial kinetic energy - Kinetic energy lost

Kinetic energy remaining = Initial kinetic energy - Potential energy gained

Kinetic energy remaining = m×78.48 - m × 9.81 × 5.4 = m×(78.48 - 52.974)

1/2×m×v₂² = m×25.506

Where:

v₂ = The speed of the block at the top of the second hill

v₂² = 2×25.506 = 51.012

v₂ = 7.14 m/s

The speed of the block at the top of the second hill = 7.14 m/s.

answera compound;

the ima (higher than 1m) will be greater than the ima with the height of 1m.