Final answer:
The diver's velocity when she lands in the pool is approximately -15.92 m/s, directed downward.
Step-by-step explanation:
To determine the diver's velocity when she lands in the pool, we need to consider her initial velocity when she jumps off the platform and her acceleration as she falls. We can use the kinematic equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is acceleration, and t is time.
Since the velocity is directed upward when she jumps off the platform, the initial velocity (u) is positive 2.6 m/s. The acceleration (a) is downward at a rate of 9.8 m/s². The time (t) can be determined using the formula:
h = ut + (1/2)at^2
where h is the height of the platform, u is the initial velocity, a is acceleration, and t is time.
In this case, h is 27 m, u is 2.6 m/s, and a is -9.8 m/s². Solving for t, we get:
t = sqrt((2h) / a) = sqrt((2 * 27) / 9.8) ≈ 1.83 seconds
Substituting the values into the velocity equation, we get:
v = 2.6 + (-9.8) * 1.83 = -15.92 m/s
Therefore, the diver's velocity the moment she lands in the pool is approximately -15.92 m/s, directed downward.