The velocity of a diver just before hitting the water is -8.8 m/s, where the minus sign indicates that her motion is directly downward. What is her displacement during the last 0.93 s of the dive?
To find the displacement of the diver during the last 0.93 s of the dive, we need to use the equation of motion relating displacement, initial velocity, time, and acceleration.
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).
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)