From the top of a cliff overlooking a lake, a person throws two stones, as shown in the drawing. The cliff is 39.2 m high. The two stones described have identical initial speeds of v0 = 19.2 m/s and are thrown at an angle θ = 29.2 °, one below the horizontal and one above the horizontal. What is the distance between the points where the stones strike the water? Neglect air resistance.

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To find the distance between the points where the stones strike the water, we can use the equations of projectile motion. Let's break down the problem step by step.

Step 1: Analyzing the motion
In this situation, we have two projectiles thrown at different angles but with the same initial speed and from the same height. One stone is thrown below the horizontal, and the other stone is thrown above the horizontal. We need to find the horizontal distance traveled by each stone until it hits the water.

Step 2: Breaking the initial velocity into components
Since we have the initial speed and angle for each stone, we can break down the initial velocity into its horizontal and vertical components. The horizontal component (vx) remains constant throughout the motion, while the vertical component (vy) changes due to the effect of gravity.

For each stone:
v0x = v0 * cos(θ)
v0y = v0 * sin(θ)

Step 3: Finding the time of flight
The time of flight (t) for each stone can be calculated using the following equation:
t = (2 * v0y) / g

Where g is the acceleration due to gravity (approximately 9.8 m/s²).

Step 4: Finding the horizontal distance traveled
Since the horizontal component of velocity remains constant throughout the motion, we can calculate the horizontal distance traveled (dx) for each stone using the following equation:
dx = v0x * t

Step 5: Calculating the distance between the points of impact
Now that we have determined the horizontal distances traveled by each stone, we can find the distance between their points of impact by subtracting the two distances:
Distance = |dx1 - dx2|

Substituting all the values into the equations will give you the final answer.

Note: Since air resistance is neglected in this problem, we assume that there is no horizontal displacement due to the effect of air drag.