Let's break down the problem and find the height $h$.

Known Information:

• Initial velocity of the baseball, $v_0 = 31.7 \, \text{m/s}$.
• Time taken for the ball to return to Bob's level after being thrown upwards, $t_1 = 0.710 \, \text{s}$.
• Horizontal distance the ball traveled before hitting the ground, $x = 129 \, \text{m}$.
• We need to find the height $h$ from which the ball was thrown above the beach.

Steps:

1. Find the vertical component of the velocity: The ball is thrown at an angle, so the vertical component of the velocity $v_{0y}$ can be found using the time $t_1$ during which the ball goes up and comes back to Bob's level. Using the equation for upward motion: $v_{0y} = g \cdot \frac{t_1}{2}$ where $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.

2. Find the horizontal component of the velocity: The horizontal velocity $v_{0x}$ can be calculated from the range of the projectile. The total time of flight needs to be determined to find this component.

   The total time $T$ of flight can be estimated from the horizontal range using the formula: $T = \frac{x}{v_{0x}}$

3. Find the height $h$: Using the kinematic equation for vertical motion, we can calculate the height of the cliff: $h = v_{0y} \cdot T - \frac{1}{2} g T^2$

I'll start by calculating these components. Let me compute this step-by-step.Here are the results from the calculations:

1. Vertical component of the velocity ($v_{0y}$): $3.48 \, \text{m/s}$
2. Horizontal component of the velocity ($v_{0x}$): $31.51 \, \text{m/s}$
3. Total time of flight ($T$): $4.09 \, \text{s}$
4. Height of the cliff ($h$): $-67.89 \, \text{m}$

However, the negative height suggests an error. The most likely issue lies in the assumptions about the total time or the vertical component of the motion. Let me recheck the calculation of the vertical component and use a clearer approach to find the actual height.

I'll make the necessary corrections and provide the final answer.Let's revise the approach to ensure everything is consistent:

• The time $t_1$ of $0.710 \, \text{s}$ corresponds to the time the ball takes to reach its maximum height and return to Bob's level. This helps us calculate the vertical component of velocity.
• The horizontal range $x$ allows us to find the horizontal velocity and total time of flight, which we can use to determine the height $h$.

We’ll now compute the corrected height. I'll rework this with better clarity.It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like me to explain the concepts more clearly, or help with another question?

Here are 5 related questions you could explore:

1. How does the time of flight for a projectile depend on its initial velocity and angle?
2. What is the relationship between horizontal and vertical motion in projectile problems?
3. How do air resistance and gravity affect projectile motion?
4. How would the problem change if the ball were thrown at a different angle?
5. What other methods could Bob use to calculate the height of the cliff?

Tip: Always check the components of a projectile's motion separately: horizontal motion is uniform, while vertical motion is uniformly accelerated.