1. Set up the condition for losing contact

A block slides on the outside of a smooth hemisphere, starting from rest at the top.

It loses contact when the normal reaction becomes zero:

$N = 0$

At an angular position $\theta$ from the top, the radial equation toward the center is

$mg\cos\theta - N = \frac{mv^2}{R}$

At the point of losing contact, $N=0$, so

$mg\cos\theta = \frac{mv^2}{R}$

or

$v^2 = gR\cos\theta \qquad (1)$

2. Apply conservation of mechanical energy

Initially, at the top, the block is at rest.

If it has moved to angle $\theta$, its vertical drop from the top is

$h = R - R\cos\theta = R(1-\cos\theta)$

By energy conservation:

$mgh = \frac12 mv^2$

Substitute $h = R(1-\cos\theta)$:

$mgR(1-\cos\theta) = \frac12 mv^2$

$v^2 = 2gR(1-\cos\theta) \qquad (2)$

3. Use equations (1) and (2)

From (1) and (2):

$gR\cos\theta = 2gR(1-\cos\theta)$

Cancel $gR$:

$\cos\theta = 2(1-\cos\theta)$

$\cos\theta = 2 - 2\cos\theta$

$3\cos\theta = 2$

$\cos\theta = \frac23$

4. Find the height fallen from the top

$h = R(1-\cos\theta)$

Given $R=3\,\text{m}$:

$h = 3\left(1-\frac23\right) = 3\cdot \frac13 = 1\,\text{m}$

So the block loses contact after descending by

$\boxed{1\,\text{m}}$

5. Compare with stored answer

Stored correct answer: $2$

Our derived answer is $1$.

This mismatch likely comes from interpreting $h$ differently. If the figure defines $h$ as the height of the block above the ground/base at the separation point, then:

$h = R\cos\theta = 3\cdot \frac23 = 2\,\text{m}$

So:

• Height descended from the top $= 1\,\text{m}$
• Height above the center/base level $= 2\,\text{m}$

Since the stored answer is $2$, the figure likely asks for the latter height.