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Projectile motion is the path that an object follows when it is thrown into the air and influenced only by gravity and its initial thrust. In simple terms, we can imagine any projectile motion as having two components: horizontal and vertical. However, in this specific exercise, we only need to consider vertical motion since the rock was thrown straight up, which simplifies calculations.

Here's how it works:

• The object, in this case, a rock, starts with an initial velocity when thrown upward.
• It ascends, constantly slowing down, until its velocity is 0 at the peak of its path.
• From the peak, it begins its descent, speeding up toward its original point and then continuing past it to the final destination.

This continual shift from moving up to down forms a parabolic path typical in projectile motion, although less curved when considering only vertical motion as with the rock. Understanding this path helps us clarify how the rock behaves and what factors play a role in determining its speed at various points, like just before hitting the water.

Free fall refers to the motion of an object being pulled down by gravity alone, without any other forces, like air resistance, acting upon it. During free fall, the object experiences constant acceleration due to gravity. The standard gravitational pull is denoted as 9.81 m/s² on Earth.

When the rock descends back past the thrower, it effectively enters free fall, with gravity solely driving its motion. Here’s what happens:

• The rock accelerates as it falls and picks up speed rapidly.
• At this stage, its initial velocity at phase two is zero because it reverses direction at the peak.
• Calculating how fast it will go at specific moments requires understanding free fall dynamics.

This insight is crucial for predicting the rock's velocity at the point of interest—just before it plunges into the river. In practice, by understanding free fall, we can apply the relevant kinematic equations to find the speed efficiently.

Kinematic equations are the fundamental tools used in physics to predict various aspects of an object's motion under constant acceleration. For this scenario, we use these equations to find out how fast the rock is moving just before contact with the water. The specific equation is:\[ v^2 = u^2 + 2as \]Here:

• \( u \) is the initial velocity of the rock when phase 2 starts, which is 0 m/s as it has reversed its direction.
• \( v \) is the final velocity that we are solving for.
• \( a \) is the acceleration from gravity, 9.81 m/s².
• \( s \) is the vertical displacement, which is 28.0 meters in this case.

Utilizing this formula, we rearrange it to solve for \( v \):\[ v^2 = 0 + 2(9.81)(28.0) = 549.36 \]By finding the square root, \( v \) is approximately 23.43 m/s.

This powerful tool helps us neatly determine the rock's velocity when facing gravitational forces, simplifying more complex motions into easily comprehensible calculations.