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
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.
To solve the problem, we will go through the steps one by one.
Step 1: Derive the time T from (1/2) g T^2 = 8.9 m
Let's solve the equation for T:
(1/2) g T^2 = 8.9 m
Multiply both sides of the equation by 2 to eliminate the fraction:
g T^2 = 2 * 8.9 m
Simplify the equation:
g T^2 = 17.8 m
Divide both sides of the equation by g:
T^2 = 17.8 m / g
Take the square root of both sides to solve for T:
T = sqrt(17.8 m / g)
Step 2: Multiply the time T required to fall 8.9 m (in seconds) by the rotation rate in rev/s.
Given the average angular speed is 1.4 rev/s, we need to calculate the number of revolutions made during the time T.
Number of Revolutions = T * Angular Speed
Number of Revolutions = sqrt(17.8 m / g) * 1.4 rev/s
Now, substitute the value of g (acceleration due to gravity) as approximately 9.8 m/s^2:
Number of Revolutions = sqrt(17.8 m / 9.8 m/s^2) * 1.4 rev/s
Number of Revolutions = sqrt(1.81) * 1.4 rev/s
Number of Revolutions = 1.346 * 1.4 rev/s
Number of Revolutions ≈ 1.88 rev
Therefore, the diver makes approximately 1.88 revolutions while on the way down.
To solve this problem, we can follow the given steps:
Step 1: Derive the time T from the equation (1/2)gT^2 = 8.9 m.
- Rewrite the equation: T^2 = (2 * 8.9 m) / g.
- Divide both sides of the equation by g: T^2 = (17.8 m) / g.
- Take the square root on both sides to solve for T: T = √[(17.8 m) / g].
Step 2: Multiply the time T required to fall 8.9 m by the rotation rate in rev/s.
- Multiply T by the given angular speed: T * 1.4 rev/s.
Step 3: Calculate the number of revolutions by simplifying the expression.
- Substitute T with √[(17.8 m) / g]: (√[(17.8 m) / g]) * 1.4 rev/s.
- Simplify the expression: 1.4 √[(17.8 m) / g] rev.
So, the number of revolutions the diver makes while on the way down is 1.4 √[(17.8 m) / g] revolutions.