## Answer :

Given:

- Mass of the ball, [tex]\( m = 0.5 \)[/tex] kg

- Height from which the ball is released, [tex]\( h = 30 \)[/tex] m

- Acceleration due to gravity, [tex]\( g = 9.8 \)[/tex] m/s²

- Initial velocity, [tex]\( u = 0 \)[/tex] m/s (since the ball is released from rest)

The relevant kinematic equation is:

[tex]\[ v^2 = u^2 + 2gh \][/tex]

Plugging in the known values:

[tex]\[ v^2 = 0 + 2 \cdot 9.8 \cdot 30 \][/tex]

First, calculate [tex]\( 2 \cdot 9.8 \cdot 30 \)[/tex]:

[tex]\[ 2 \cdot 9.8 = 19.6 \][/tex]

[tex]\[ 19.6 \cdot 30 = 588 \][/tex]

Now, we have:

[tex]\[ v^2 = 588 \][/tex]

To find [tex]\( v \)[/tex], take the square root of both sides:

[tex]\[ v = \sqrt{588} \][/tex]

When you calculate the square root of 588, you get:

[tex]\[ v \approx 24.24871130596428 \][/tex]

Therefore, the velocity of the ball when it hits the ground is approximately [tex]\( 24.2 \)[/tex] m/s.

So, the correct answer is:

[tex]\[ \boxed{24.2 \text{ m/s}} \][/tex]

Thus, the answer is option A: [tex]\(24.2 \text{ m/s}\)[/tex].