I need help solving the problem below. Please respond.

At the local swimming hole, a favorite trick is to run horizontally off a cliff that is 8.9 m above the water. One diver runs off the edge of the cliff, tucks into a "ball," and rotates on the way down with an average angular speed of 1.4 rev/s. Ignore air resistance and determine the number of revolutions she makes while on the way down.

Multiply the time T required to fall 8.9 m (in seconds) by the rotation rate in rev/s.

Derive the time T from
(1/2) g T^2 = 8.9 m

1 answer

First, let's find the time T. To do this, we will use the formula (1/2)gT^2 = 8.9m, where g is the acceleration due to gravity (approx. 9.8 m/s^2).

(1/2)(9.8 m/s^2)(T^2) = 8.9 m

Now, we solve for T^2:

T^2 = (2 * 8.9 m) / (9.8 m/s^2)

T^2 ≈ 1.8163

Now, we take the square root of both sides:

T ≈ sqrt(1.8163) ≈ 1.3486 s

Now that we have the time, we can calculate the number of revolutions the diver makes while falling. We will multiply the time (in seconds) by the rotation rate (in rev/s):

Number of revolutions = Time × Rotation rate
Number of revolutions = 1.3486 s × 1.4 rev/s

Number of revolutions ≈ 1.88804

So, the diver makes approximately 1.89 revolutions while on the way down.