well, consider the reverse: if she were going up at an initial velocity of 8.8 m/s, how high would she go in .93 sec?
h= 8.8*.93-1/2 g (.93^2)
h= 8.8*.93-1/2 g (.93^2)
Since the diver is moving directly downward, we can assume her acceleration due to gravity is 9.8 m/s^2.
The equation of motion is given as:
displacement = initial velocity * time + (1/2) * acceleration * time^2
Given:
Initial velocity (u) = -8.8 m/s (negative sign indicates downward motion)
Time (t) = 0.93 s
Acceleration (a) = 9.8 m/s^2 (assuming downward)
Plugging in the values into the equation of motion:
displacement = (-8.8 m/s) * (0.93 s) + (1/2) * (9.8 m/s^2) * (0.93 s)^2
Calculating the displacement:
displacement = -8.184 m + (1/2) * 9.8 m/s^2 * 0.8649 s^2
displacement = -8.184 m + 4.268 m
displacement = -3.916 m
Therefore, the displacement of the diver during the last 0.93 s of the dive is approximately -3.92 m. The negative sign indicates that her final position is below her initial position (as she is moving downward).