91Ó°ÊÓ

Horizontal velocity refers to the component of velocity parallel to the earth's surface. When an object leaves a point and moves horizontally, as in the case of the swimmer on the slide, their horizontal velocity remains constant throughout the motion. This is because there are no forces acting on the object in the horizontal direction, such as air resistance, which we are neglecting in this problem.

To calculate horizontal velocity (\(v_x\)), we use the simple formula:

• \(d_x = v_x \cdot t\)

Where \(d_x\) is the horizontal distance traveled, and \(t\) is the time taken. For our water slide problem, the swimmer travels a distance of \(5.00 \text{ m}\) in \(0.500 \text{ s}\), giving them a constant horizontal velocity of \(10.0 \text{ m/s}\). This uniform motion condition helps simplify the separation of horizontal and vertical aspects of projectile motion.

Vertical motion is quite different from horizontal motion because it is influenced by gravity. When an object falls, it accelerates vertically, even if it starts from rest, due to gravitational pull. For the swimmer, this aspect of motion starts immediately after leaving the slide.

In the case of vertical projectile motion, the swimmer's initial vertical velocity (\(v_{y0}\)) is zero because they are "launched" horizontally. The height they fall through (\(H\)) can be determined using the formula:

• \(H = \frac{1}{2} g t^2\)

With \(g\) being acceleration due to gravity (\(9.81 \text{ m/s}^2\)) and \(t\) being the time they took to hit the water (\(0.500 \text{ s}\)). This formula arises from the equations of motion, considering the initial vertical velocity is zero.

Gravity is a constant acceleration that affects all objects in free fall. In our context, its value is \(9.81 \text{ m/s}^2\). This acceleration continuously influences the object's vertical velocity and causes the swimmer to descend towards the water.

When solving projectile motion problems, gravity is a key factor in computing how long an object will remain in the air, and how far it will fall. For objects moving horizontally and then dropping (like our swimmer), the uniform gravitational pull accelerates the object downwards, leading to a calculated fall height using the formula:

• \( H = \frac{1}{2} g t^2 \)

This equation shows that the fall height is dependent on both the gravitational acceleration and the square of the time. It vividly illustrates how quickly distances can increase with a passing of time even with short durations like \(0.500 \text{ s}\).