To determine the speed of the ball when it hits the ground after being released from a height of 30 meters, we need to use the principles of physics, specifically the equations of motion under constant acceleration due to gravity.
Given:
- Mass of the ball, [tex]\( m = 0.5 \, \text{kg} \)[/tex]
- Height from which the ball is released, [tex]\( h = 30 \, \text{m} \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
- Initial velocity, [tex]\( u = 0 \, \text{m/s} \)[/tex] (the ball is released from rest)
The formula we use to find the final velocity ([tex]\( v \)[/tex]) of an object under free fall is derived from the kinematic equation:
[tex]\[ v^2 = u^2 + 2gh \][/tex]
Since the initial velocity [tex]\( u \)[/tex] is 0, this simplifies to:
[tex]\[ v^2 = 2gh \][/tex]
Now, substituting the given values for [tex]\( g \)[/tex] and [tex]\( h \)[/tex]:
[tex]\[ v^2 = 2 \times 9.8 \, \text{m/s}^2 \times 30 \, \text{m} \][/tex]
[tex]\[ v^2 = 588 \, \text{m}^2/\text{s}^2 \][/tex]
To find the final velocity [tex]\( v \)[/tex], we take the square root of both sides:
[tex]\[ v = \sqrt{588 \, \text{m}^2/\text{s}^2} \][/tex]
[tex]\[ v \approx 24.24871130596428 \, \text{m/s} \][/tex]
Thus, after considering the precision and rounding to one decimal place, the final velocity of the ball when it hits the ground is approximately [tex]\( 24.2 \, \text{m/s} \)[/tex].
So, the correct answer is:
[tex]\[ \boxed{24.2 \, \text{m/s}} \][/tex]
Therefore, the answer is:
B. [tex]\( 24.2 \, \text{m/s} \)[/tex]
