91Ó°ÊÓ

The concept of horizontal velocity in projectile motion is straightforward. When an object moves horizontally off an edge, just like the tennis ball in our exercise, it maintains a constant horizontal velocity until it hits the ground. This is because no horizontal forces act on the ball, as air resistance is negligible in these calculations.

In our tennis ball scenario, to find the horizontal velocity, we use the equation:

• \( x = v_{0x} \times t \)

Here, \( x \) is the horizontal distance traveled (1.40 meters), and \( t \) is the time of flight of the ball (0.391 seconds).
By rearranging the equation, we get:

• \( v_{0x} = \frac{x}{t} = \frac{1.40}{0.391} \approx 3.58 \text{ m/s} \)

This calculation shows the ball's horizontal velocity as 3.58 m/s without any change during its flight.

Vertical velocity is a crucial component in understanding projectile motion. Unlike horizontal velocity, the vertical velocity of a projectile changes due to gravity's effects. Initially, when our tennis ball leaves the table, its vertical velocity is zero because it starts from rest in the vertical direction.

As the ball falls, it accelerates downward at \(9.81 \text{ m/s}^2\) (the acceleration due to gravity). The formula to calculate the vertical velocity just before it strikes the ground is:

• \( v_{y} = g \times t \)

Using our time of flight \( t = 0.391 \text{ seconds} \), the calculation becomes:

• \( v_{y} = 9.81 \times 0.391 \approx 3.84 \text{ m/s} \)

Thus, by the time the tennis ball hits the ground, it reaches a vertical velocity of 3.84 m/s.

In projectile motion, the time of flight is the total time the object is in the air. To solve for this, especially when the initial vertical motion isn't provided, we utilize the equation for free fall:

• \( y = \frac{1}{2} g t^2 \)

Where \( y \) is the vertical drop (0.750 meters) and \( g \) is the acceleration due to gravity \(9.81 \text{m/s}^2\).
Solving for \( t \) involves:

• \( t = \sqrt{\frac{2 \times 0.750}{9.81}} \approx 0.391 \text{ seconds} \)

Thus, from edge to impact, the tennis ball is airborne for approximately 0.391 seconds.

The magnitude of the velocity is a measure of how fast an object is moving just before it impacts the ground. For an object in projectile motion, this is calculated by considering both the horizontal and vertical components of velocity.

The formula used to find this overall velocity just before impact is given by the Pythagorean theorem:

• \( v = \sqrt{v_{0x}^2 + v_{y}^2} \)

Where \( v_{0x} = 3.58 \text{ m/s} \) (horizontal) and \( v_{y} = 3.84 \text{ m/s} \) (vertical).
Calculate the magnitude as:

• \( v = \sqrt{3.58^2 + 3.84^2} \approx 5.25 \text{ m/s} \)

This means the ball's total speed, right before hitting the floor, is about 5.25 m/s.

Understanding both the magnitude and direction of velocity gives us a complete picture of the object’s speed and path. The direction at which the tennis ball hits the floor is given by the angle \( \theta \) below the horizontal.

This angle can be determined using the arctangent of the ratio of vertical to horizontal velocity:

• \( \theta = \arctan\left(\frac{v_{y}}{v_{0x}}\right) \)

Substitute the known values:

• \( \theta = \arctan\left(\frac{3.84}{3.58}\right) \approx 46.5^\circ \)

Therefore, the ball strikes the floor at an angle of approximately 46.5° below the horizontal, highlighting the path's steepness right before impact.