To determine the speed at which the ball hits the ground, we can use the principle of conservation of energy. Specifically, the potential energy of the ball at the height will be converted into kinetic energy as it falls.
1. Potential energy at the top:
- The potential energy ([tex]\(PE\)[/tex]) at a height ([tex]\(h\)[/tex]) is given by the formula [tex]\(PE = mgh\)[/tex], where:
- [tex]\(m\)[/tex] is the mass of the ball.
- [tex]\(g\)[/tex] is the acceleration due to gravity.
- [tex]\(h\)[/tex] is the height.
Given:
- [tex]\(m = 0.3 \, \text{kg}\)[/tex]
- [tex]\(h = 8 \, \text{m}\)[/tex]
- [tex]\(g = 9.8 \, \text{m/s}^2\)[/tex]
So,
[tex]\[
PE = 0.3 \times 9.8 \times 8 = 23.52 \, \text{J} \quad (\text{Joules})
\][/tex]
2. Kinetic energy at the bottom:
- The kinetic energy ([tex]\(KE\)[/tex]) at the bottom is given by the formula [tex]\(KE = \frac{1}{2} mv^2\)[/tex], where:
- [tex]\(v\)[/tex] is the speed of the ball.
At the moment the ball hits the ground, all the potential energy will have been converted to kinetic energy. Therefore, [tex]\(PE = KE\)[/tex]:
[tex]\[
PE = \frac{1}{2} mv^2 = 23.52 \, \text{J}
\][/tex]
3. Solving for [tex]\(v\)[/tex]:
[tex]\[
23.52 = \frac{1}{2} \times 0.3 \times v^2
\][/tex]
First, we simplify the equation:
[tex]\[
23.52 = 0.15 \times v^2
\][/tex]
Then, solve for [tex]\(v^2\)[/tex]:
[tex]\[
v^2 = \frac{23.52}{0.15} = 156.8
\][/tex]
Finally, take the square root of both sides to find [tex]\(v\)[/tex]:
[tex]\[
v = \sqrt{156.8} \approx 12.521980673998822 \, \text{m/s}
\][/tex]
So the speed of the ball when it hits the ground is approximately [tex]\(12.5 \, \text{m/s}\)[/tex].
Answer: A. [tex]\(12.5 \, \text{m/s}\)[/tex]
