To determine how far Dave Johnson's center of mass moves up during the high jump, we can use the principles of projectile motion. Here's how we can calculate it:
1. Given information:
- Initial vertical velocity (v0) = 5.00 m/s
- Acceleration due to gravity (g) = 9.8 m/s² (assuming Earth's gravity)
- Final vertical velocity (vf) = 0 m/s (at the highest point of the jump)
2. The motion of Dave Johnson's center of mass can be treated as a projectile, with an initial upward velocity and downward acceleration due to gravity.
3. At the highest point of the jump, the final vertical velocity (vf) is 0 m/s. This is because the velocity changes direction from upward to downward due to gravity.
4. We can use the equation vf = v0 + gt, where t is the time taken to reach the highest point.
5. Solving for t:
- 0 = 5.00 m/s + (9.8 m/s²)t
- -5.00 m/s = (9.8 m/s²)t
- t = -0.51 s (rounded to two decimal places)
Note: The negative sign indicates that the initial upward velocity and the downward acceleration due to gravity cancel each other out at the highest point.
6. To find the distance the center of mass moves up, we can use the kinematic equation s = v0t + (1/2)gt².
7. Substituting the known values:
- s = (5.00 m/s)(-0.51 s) + (1/2)(9.8 m/s²)(-0.51 s)²
- s = -2.55 m - 0.653 m
- s = -3.20 m (rounded to two decimal places)
8. The distance the center of mass moves up during the high jump is 3.20 meters.
Note: The negative sign indicates that the center of mass moves in the opposite direction of the initial velocity, which is downward in this case.
In summary, the center of mass moves up 3.20 meters during Dave Johnson's high jump.