Question

A racquet ball is struck in such a way that it leaves the racquet with a speed of \(4.87 \mathrm{~m} / \mathrm{s}\) in the horizontal direction. When the ball hits the court, it is a horizontal distance of \(1.95 \mathrm{~m}\) from the racquet. Find the height of the racquet ball when it left the racquet.

Short answer

The height is approximately 0.785 meters.

Step-by-step solution

Understand the Problem

We have a ball with an initial speed of 4.87 m/s launched horizontally. It travels 1.95 m horizontally before hitting the ground. We need to find the vertical height from which it was launched.

Determine Time of Flight

Since the ball travels horizontally at a constant speed, we can find the time it takes to hit the ground using the formula: \[ t = \frac{\text{horizontal distance}}{\text{horizontal speed}} = \frac{1.95}{4.87} \approx 0.4 \text{ seconds} \]

Use Vertical Motion Formula

For the vertical motion, we use the formula for free fall under gravity: \[ h = \frac{1}{2} g t^2 \] where \( g = 9.81 \text{ m/s}^2 \) and \( t \approx 0.4 \text{ seconds} \).

Calculate the Height

Substitute the values into the formula: \[ h = \frac{1}{2} \times 9.81 \times (0.4)^2 = \frac{1}{2} \times 9.81 \times 0.16 \approx 0.7848 \text{ meters} \]

Conclusion

The height of the racquetball when it left the racquet is approximately 0.785 meters above the ground.

Key concepts

Horizontal Velocity

When we talk about projectile motion, horizontal velocity is a key concept. In this scenario, the racquetball has been struck with an initial velocity of 4.87 meters per second. It's important to recognize that horizontal velocity remains constant throughout the motion of the ball, because there are no external forces going sideways (like air resistance) acting on it. This constancy is because velocity in the horizontal direction is not affected by gravity. So, when a ball is launched horizontally, you'll find it maintains the same horizontal speed until it hits the ground.
This simplification helps in computing other parameters such as the time of flight, making it easier to assess projectile paths.

Time of Flight

Time of flight refers to the duration for which the ball is in the air. For a horizontally launched projectile, determining this time is a straight-forward calculation. Since the horizontal velocity is constant, you can use the formula:

• \( t = \frac{\text{horizontal distance}}{\text{horizontal speed}} \)

This formula simply divides the total horizontal distance by the horizontal speed. In our problem, the ball's horizontal distance traveled is 1.95 meters and the speed is 4.87 meters per second. Therefore, using the formula:

• \( t = \frac{1.95}{4.87} \approx 0.4 \text{ seconds} \)

This time then helps in finding out how far the ball falls vertically during the flight.

Free Fall Formula

In studying the vertical motion of a projectile, the 'free fall formula' is indispensable. The vertical component of motion is influenced by gravity. When something falls freely under gravity, its vertical distance fallen can be calculated using the formula:

• \( h = \frac{1}{2} g t^2 \)

where \( h \) is the height, \( g \) is the acceleration due to gravity (approximately 9.81 \( \text{m/s}^2 \) on Earth), and \( t \) is the time of flight.
This scenario's time of flight \( t \) is approximately 0.4 seconds. Thus, substituting these values into our formula, we get:

• \( h = \frac{1}{2} \times 9.81 \times (0.4)^2 \approx 0.7848 \text{ meters} \)

This calculation shows the ball's initial vertical height when it left the racquet, reflecting the height from which it initially started its motion.