A ball of mass 0.5 kg is released from rest at a height of 30 m. How fast is it going when it hits the ground? (Acceleration due to gravity is [tex]\( g = 9.8 \, m/s^2 \)[/tex].)

A. [tex]\( 24.2 \, m/s \)[/tex]
B. [tex]\( 31.2 \, m/s \)[/tex]
C. [tex]\( 8.6 \, m/s \)[/tex]
D. [tex]\( 3.1 \, m/s \)[/tex]

Answer :

To determine how fast the ball is going when it hits the ground, we will use the principles of conservation of energy or kinematic equations for motion under gravity. Here, we'll use the kinematic equation for motion under gravity.

- 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].

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