Final answer:
Using the kinematic equation for uniformly accelerated motion, it is found that the gymnast's center of mass will be approaching the apex of her back flip trajectory 1 second after take-off with an initial vertical velocity of 9.81 m/s.
Step-by-step explanation:
If a gymnast leaves the ground with a vertical velocity of 9.81 m/s when doing a back flip, the point in her trajectory where her center of mass will be after 1.0 sec can be determined by using the kinematic equations for uniformly accelerated motion. Because the only acceleration acting on the gymnast is the acceleration due to gravity (g = 9.81 m/s2), we can use the following kinematic equation to find the displacement (s) after time (t):
s = ut + 1/2 at2
where:
- u is the initial vertical velocity (9.81 m/s)
- t is the time (1.0 s)
- a is the acceleration (which is -g, since gravity is acting downwards)
Plugging in the values gives us:
s = (9.81 m/s)(1.0 s) + 1/2 (-9.81 m/s2)(1.0 s)2
s = 9.81 m - 4.905 m
s = 4.905 m
This indicates that after 1 second, the gymnast's center of mass is still ascending and has not yet reached the apex of the flip, which corresponds to the highest point. Therefore, the correct answer is option (a) Approaching the apex.