Question

A ball is launched with initial speed v from the ground level up a frictionless hill. The hill becomes steeper as the ball slides up; however, the ball remains in contact with the hill at all times. Using conservation of energy, find the maximum vertical height hmax to which the ball will climb.

Answer 1

Answer:

H(max) = (v²/2g)

Explanation:

The maximum height the ball will climb will be when there is no friction at all on the surface of the hill.

Normally, the conservation of kinetic energy (specifically, the work-energy theorem) states that, the change in kinetic energy of a body between two points is equal to the work done in moving the body between the two points.

With no frictional force to do work, all of the initial kinetic emergy is used to climb to the maximum height.

ΔK.E = W

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Final kinetic energy = 0 J, (since the body comes to rest at the height reached)

Initial kinetic energy = (1/2)(m)(v²)

Workdone in moving the body up to the height is done by gravity

W = - mgH

ΔK.E = W

0 - (1/2)(m)(v²) = - mgH

mgH = mv²/2

gH = v²/2

H = v²/2g.