Question

An Olympic diver performs a dive from a platform 27 m above the surface of the pool. The diver

jumps off the platform with a velocity of 2.6 m/s upward and then accelerates at a constant rate
of 9.8 m/s2 downward as she falls. What is the diver's velocity the moment she lands in the pool
below?

Answer 1

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.