An object is at rest on top of a smooth sphere with a radius of? = 15.3 m that is buried exactly halfway under the ground. It then begins to slide down. At what height from the ground is the object no longer in contact with the sphere?

10.2 m.

Explanation:

Let object falls by angle θ.

At any moment after the fall, there are two forces acting on the sphere

1) mg cosθ which is a component of weight towards the centre 2) normal reaction of the surface R.

mgcosθ - R is net force acting, which provides centripetal force

mgcosθ - R = mv² / r

But v² = 2g r (1-cosθ) [ object falls by height (r - r cosθ).

mgcosθ - R = m / r x 2g r (1-cosθ)

When the object is no longer in touch with sphere,

R = 0

mgcosθ = m / r x 2g r (1-cosθ)

3 gr cosθ = 2gr

cosθ = 2/3

height of fall

= r (1-cosθ)

r (1 - 2/3)

1/3 r

1/3 x 15.3

5.1 m

Height from the ground

15.3 - 5.1

10.2 m.