Final answer:
The shell lands approximately 30.9 m past the edge of the cliff.
Step-by-step explanation:
To find the horizontal distance the shell lands past the edge of the cliff, we can break down the motion into vertical and horizontal components. Since the shell is launched at an angle of 43.0° above the horizontal, we need to find the time it takes for the shell to reach the cliff height, then use that time to calculate the horizontal distance.
First, let's find the time it takes for the shell to reach the cliff height. We can use the vertical motion equation:
Final vertical displacement = Initial vertical velocity * Time + 0.5 * Acceleration * Time^2
Since the final vertical displacement is 25.0 m (the height of the cliff) and the initial vertical velocity is 32.6 m/s (sin(43.0°) * 32.6 m/s), we can rearrange the equation to solve for time:
Time = (Final vertical displacement - 0.5 * Acceleration * Time^2) / (Initial vertical velocity)
Plugging in the values:
Time = (25.0 m - 0.5 * (-9.8 m/s^2) * Time^2) / (32.6 m/s * sin(43.0°))
Time = 1.15 s
Now that we have the time it takes to reach the cliff height, we can use this time to calculate the horizontal distance. The horizontal distance can be found using the horizontal motion equation:
Horizontal distance = Initial horizontal velocity * Time
Since the initial horizontal velocity is 32.6 m/s (cos(43.0°) * 32.6 m/s) and the time is 1.15 s, we can calculate the horizontal distance:
Horizontal distance = 32.6 m/s * cos(43.0°) * 1.15 s
Horizontal distance = 30.9 m.