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This is a physics problem (Problem 13.83) about a cliff diver in Acapulco. Let's break down the details from the image:

### Given:
- The mean water depth \( h = 4 \, \text{m} \).
- The crests of the waves are 1 meter above the mean water depth, i.e., the crest height is \( 5 \, \text{m} \) from the bottom of the sea.
- The horizontal velocity of the waves is \( \sqrt{gh} \), where \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).
- The diver’s aiming point is 2 meters horizontally from the base of the cliff.
- The diver's velocity is horizontal when he begins his dive.

### Questions:
**(a)** What is the magnitude of the diver's velocity when he enters the water?  
**(b)** How far from his aiming point must a wave crest be when he dives in order for him to enter the water at the crest?

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### Solution Steps:

#### Part (a): The diver's velocity when entering the water
The key here is that the diver's motion involves free fall vertically due to gravity. Assuming his initial vertical velocity is zero (since he begins with horizontal velocity only), we can calculate his vertical velocity just before hitting the water.

1. **Vertical displacement**: 
   The diver must fall from a height of 5 meters (since the wave crests are 1 meter above the mean water depth of 4 meters).

   The time \( t \) for free fall can be calculated using:
   \[
   h = \frac{1}{2} g t^2
   \]
   Solving for \( t \):
   \[
   t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5}{9.81}} \, \text{s}
   \]

2. **Horizontal velocity**: 
   The problem states that the diver begins with horizontal velocity and does not decelerate horizontally. We are not given the diver’s initial horizontal velocity directly, but the horizontal velocity remains constant throughout the dive.

3. **Total velocity**: 
   The diver's total velocity when entering the water is the vector sum of the vertical and horizontal components.

#### Part (b): Distance from the aiming point for wave crest synchronization
The diver and the wave need to meet at the same point when the diver enters the water. The waves travel horizontally with velocity \( \sqrt{gh} \). Using the time \( t \) from part (a), you can calculate how far the wave crest must be from the diver’s aiming point for him to meet it at the water's surface.

Let me first calculate the relevant values for part (a), starting with \( t \). I'll get those calculations underway.The time for the diver to fall 5 meters is approximately \( 1.01 \, \text{seconds} \).

### Now, let's proceed with part (a):

To find the total velocity when the diver enters the water, we need:

1. The **vertical velocity** right before hitting the water, \( v_y = g \cdot t \).
2. The **horizontal velocity** remains constant throughout the dive, but since it's not provided directly, we might focus on the vertical velocity for now or leave the horizontal velocity symbolic.

I'll calculate the vertical velocity \( v_y \) next.The vertical velocity of the diver just before hitting the water is approximately \( 9.90 \, \text{m/s} \).

To fully answer part (a), we would also need the horizontal velocity, but since it's not given, the magnitude of the diver's velocity upon entering the water would be the vector sum of this vertical velocity and his horizontal velocity.

### Part (b):
The horizontal velocity of the wave is given by \( \sqrt{gh} \). Using the free fall time \( t \), we can calculate how far the wave crest must be from the diver's aiming point at the time of the dive.

I'll calculate this distance now.The wave must be approximately \( 6.32 \, \text{m} \) away from the diver's aiming point when he dives, in order for him to enter the water at the crest.

### Summary:
- **(a)** The vertical velocity of the diver when entering the water is approximately \( 9.90 \, \text{m/s} \). The total velocity would depend on the horizontal component, which is not provided.
- **(b)** The wave crest must be \( 6.32 \, \text{m} \) away from the aiming point when the diver begins his dive.

Would you like to explore the problem further, or have any specific questions?

### Related questions:
1. What is the total velocity if the horizontal velocity is given or assumed?
2. How does the wave speed compare to the diver's horizontal velocity?
3. What if the diver jumped with an initial vertical velocity?
4. How would the problem change if the wave velocity were slower?
5. Can we calculate the horizontal velocity of the diver if more data were provided?

### Tip:
Always double-check if you're dealing with horizontal and vertical components separately in physics problems involving motion in two dimensions.

The cliff divers of Acapulco, Mexico must time their dives so that they enter the water at the crest of a wave. The crests of the waves are 1 m above the mean water depth h = 4 m. The horizontal velocity of the waves is equal to √gh. The diver’s aiming point is 2 m out from the base of the cliff. Assume that his velocity is horizontal when he begins the dive. (a) What is the magnitude of the diver's velocity when he enters the water? (b) How far from his aiming point must a wave crest be when he dives in order for him to enter the water at the crest?

Explore how a cliff diver times his dive to enter the water at the crest of a wave, using kinematic equations for free fall and wave motion. Solve for the diver's velocity and the wave crest position.

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