Using conservation of energy, find the speed vb of the block at the bottom of the ramp. express your answer in terms of some or all the variables m, v, and h and any appropriate constants.

√(v² + 2gh)

Explanation:

PE₀ + KE₀ = PEƒ + KEƒ

At the top of the ramp, potential energy will be at a peak and kinetic energy will be at a least. At the bottom of the ramp, kinetic energy will be at a peak and potential energy will be least . Although the type of energy changes, the total amount of energy remains the same. Meaning energy can neither be created nor destroyed, but can be transformed from one point to another.

mgh + ½mv² = mgh(b) + ½mv(b)²

Since the height at the bottom is zero, potential energy can be removed from the right side of the formula:

mgh + ½mv² = ½mv(b)²

gh + ½v² = ½v(b)²

2gh + v² = v(b)²

v(b) = √(v² + 2gh)

= √(v² + 2gh)

i belive it is d. both a and c

hope this

mark me branliest, only got   it 1

show the diagram then i will answer