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This is a physics problem that can be solved using principles from kinematics. Let's break it down step by step:

### Known values:
- Initial velocity of the ball: \( v_0 = 33.3 \, \text{m/s} \)
- Time to return to the level of Bob: \( t_1 = 0.510 \, \text{s} \)
- Horizontal distance to the base of the cliff: \( x = 125 \, \text{m} \)

The goal is to determine the height of the cliff.

### Step 1: Analyze vertical motion
Since Bob observes that the ball returns to his level after \( t_1 = 0.510 \, \text{s} \), this time represents the time it took for the ball to rise and fall back to Bob's height. The vertical velocity at this point would be zero. 

Using the equation for vertical motion under constant acceleration due to gravity \( g = 9.81 \, \text{m/s}^2 \), we can find the vertical component of the ball’s velocity.

The time to the top (where vertical velocity becomes zero) is half of the round trip, so:

\[
t_{\text{up}} = \frac{t_1}{2} = \frac{0.510}{2} = 0.255 \, \text{s}
\]

The vertical component of the initial velocity is related to time by:

\[
v_{\text{vert}} = g \cdot t_{\text{up}} = 9.81 \, \text{m/s}^2 \times 0.255 \, \text{s} = 2.5 \, \text{m/s}
\]

### Step 2: Resolve components of the initial velocity
Bob throws the ball at an angle, so we need to resolve the initial velocity into horizontal and vertical components.

Let:
- \( v_{0x} \) be the horizontal component of the velocity.
- \( v_{0y} \) be the vertical component of the velocity.

Using \( v_{0y} = 2.5 \, \text{m/s} \), we can use the Pythagorean theorem to find the horizontal component:

\[
v_{0x} = \sqrt{v_0^2 - v_{0y}^2} = \sqrt{33.3^2 - 2.5^2} \, \text{m/s}
\]

Let me calculate that now.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can calculate the horizontal velocity component using this formula:

\[
v_{0x} = \sqrt{v_0^2 - v_{0y}^2}
\]

Substitute \( v_0 = 33.3 \, \text{m/s} \) and \( v_{0y} = 2.5 \, \text{m/s} \) to find \( v_{0x} \). Once you have that, you can proceed with the next steps of solving the problem by working with the time of flight and horizontal distance traveled.

Let me know if you need help with further steps or other details!

### Related Questions:
1. What equation should be used to calculate the total time of flight for the ball?
2. How can the horizontal distance relate to the time and velocity components?
3. What assumptions are made about the motion of the ball in this type of problem?
4. How does air resistance affect real-world projectile motion compared to idealized calculations?
5. How can Bob use the measured horizontal distance to determine the height of the cliff?

### Tip:
When working on projectile motion, it helps to split the problem into horizontal and vertical components, solving each independently using kinematic equations.

Bob has just finished climbing a sheer cliff above a beach and wants to figure out how high he climbed. He uses a baseball, a stopwatch, and a friend on the ground with a long measuring tape. Bob throws the ball with an initial velocity of 33.3 m/s, and after 0.510 seconds, it is level with him again. The horizontal distance to where the ball hits the ground is 125 m from the base of the cliff. How high is the cliff?

Learn how to calculate the height of a cliff using projectile motion concepts. Given initial velocity, time, and horizontal distance, we find the height through kinematic equations. This problem demonstrates key concepts in physics, such as motion in two dimensions.

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