91Ó°ÊÓ

Kinematic equations are vital tools for understanding motion, especially in physics problems related to projectile motion like the one described in the exercise. These equations allow us to connect different aspects of motion, such as displacement, initial velocity, acceleration, and time. In this scenario, the main kinematic equation used is:

\[ v^2 = u^2 + 2as \]
This equation relates the final velocity \( v \) of an object to its initial velocity \( u \), the acceleration \( a \), and the displacement \( s \). When a projectile, like a cannonball, is launched vertically, its acceleration is typically due to gravity. Thus, \( a \) is \(-9.8 \text{ m/s}^2\), as gravity acts downward.

Understanding how to apply this equation allows us to determine one of the most critical variables in projectile problems: the initial velocity. Rearranging these terms helps make the connection between each physical quantity clear and step us toward finding solutions to real-world motion problems.

The initial velocity is the speed of an object right when it begins its motion. It's crucial because it determines how far or high an object can travel. In our exercise, we need to find the initial speed at which the cannonball was fired to just reach the top of a 61.5-meter cliff.

By setting the final velocity at the top to zero, since the cannonball stops momentarily at its peak, and substituting the known values into the kinematic equation \( 0 = u^2 + 2(-9.8)(61.5) \), we solve for \( u \). This process involves rearranging and solving the equation:

• Ensure the right side simplifies correctly to yield \( u^2 = 2 \times 9.8 \times 61.5 \).
• Take the square root to find \( u \approx 34.75 \text{ m/s} \).

This value reveals how fast the cannonball needs to be initially launched to barely reach the cliff's top, providing a basis for calculating connected aspects like range in subsequent parts of the exercise.

Horizontal range is the horizontal distance a projectile travels. It's a key insight into the motion path of objects like cannonballs. In this problem, we analyze the range from two perspectives: one cannon at the cliff's base firing upwards, and another at the top firing horizontally.

For the cannon at the base, the maximum horizontal range when fired with an upward trajectory first reaches the peak height and then falls back down. This involves calculating the time it takes to reach the peak and during its descent.

• Time to get to the highest point is \( t = \frac{u}{g} \).
• Total flight time to return to the base is about twice this value.
• The range is this time multiplied by initial speed \( u \).

For the cannon at the top, fired horizontally, the range can be found by calculating the time it takes to hit the ground, \( t = \sqrt{\frac{2h}{g}} \), and multiplying this time by the speed \( u \) itself.

Ultimately, understanding horizontal range helps us validate that some scenarios yield similar ranges despite being launched from varied elevations, demonstrating the robustness and predictability of projectile motion under constant initial speeds.