Answer :
Sure, let's determine the final speed of the table tennis ball using the principles of kinematics in physics.
Since the ball is dropped from rest, its initial velocity ([tex]\(v_i\)[/tex]) is 0 m/s.
Given:
- The height ([tex]\(h\)[/tex]) from which the ball is dropped: 13 m
- The acceleration due to gravity ([tex]\(g\)[/tex]): 9.81 m/s[tex]\(^2\)[/tex]
We can use the kinematic equation that relates initial velocity, acceleration, distance, and final velocity. The specific kinematic equation we will use is:
[tex]\[ v_f^2 = v_i^2 + 2gh \][/tex]
Where:
- [tex]\(v_f\)[/tex] is the final velocity.
- [tex]\(v_i\)[/tex] is the initial velocity.
- [tex]\(g\)[/tex] is the acceleration due to gravity.
- [tex]\(h\)[/tex] is the height from which the ball is dropped.
Step-by-step solution:
1. Initial Setup:
[tex]\[ v_i = 0 \text{ m/s} \\ h = 13 \text{ m} \\ g = 9.81 \text{ m/s}^2 \][/tex]
2. Substitute the known values into the equation:
[tex]\[ v_f^2 = 0^2 + 2 \times 9.81 \text{ m/s}^2 \times 13 \text{ m} \][/tex]
3. Calculate the product inside the equation:
[tex]\[ v_f^2 = 2 \times 9.81 \text{ m/s}^2 \times 13 \text{ m} \][/tex]
[tex]\[ v_f^2 = 255.06 \text{ m}^2/\text{s}^2 \][/tex]
4. Take the square root of both sides to solve for [tex]\(v_f\)[/tex]:
[tex]\[ v_f = \sqrt{255.06 \text{ m}^2/\text{s}^2} \][/tex]
[tex]\[ v_f \approx 15.97 \text{ m/s} \][/tex]
Therefore, the final speed of the ball, just before it reaches the ground, is approximately [tex]\(16 \text{ m/s}\)[/tex].
Since the ball is dropped from rest, its initial velocity ([tex]\(v_i\)[/tex]) is 0 m/s.
Given:
- The height ([tex]\(h\)[/tex]) from which the ball is dropped: 13 m
- The acceleration due to gravity ([tex]\(g\)[/tex]): 9.81 m/s[tex]\(^2\)[/tex]
We can use the kinematic equation that relates initial velocity, acceleration, distance, and final velocity. The specific kinematic equation we will use is:
[tex]\[ v_f^2 = v_i^2 + 2gh \][/tex]
Where:
- [tex]\(v_f\)[/tex] is the final velocity.
- [tex]\(v_i\)[/tex] is the initial velocity.
- [tex]\(g\)[/tex] is the acceleration due to gravity.
- [tex]\(h\)[/tex] is the height from which the ball is dropped.
Step-by-step solution:
1. Initial Setup:
[tex]\[ v_i = 0 \text{ m/s} \\ h = 13 \text{ m} \\ g = 9.81 \text{ m/s}^2 \][/tex]
2. Substitute the known values into the equation:
[tex]\[ v_f^2 = 0^2 + 2 \times 9.81 \text{ m/s}^2 \times 13 \text{ m} \][/tex]
3. Calculate the product inside the equation:
[tex]\[ v_f^2 = 2 \times 9.81 \text{ m/s}^2 \times 13 \text{ m} \][/tex]
[tex]\[ v_f^2 = 255.06 \text{ m}^2/\text{s}^2 \][/tex]
4. Take the square root of both sides to solve for [tex]\(v_f\)[/tex]:
[tex]\[ v_f = \sqrt{255.06 \text{ m}^2/\text{s}^2} \][/tex]
[tex]\[ v_f \approx 15.97 \text{ m/s} \][/tex]
Therefore, the final speed of the ball, just before it reaches the ground, is approximately [tex]\(16 \text{ m/s}\)[/tex].
