91Ó°ÊÓ

When a ball is thrown, the total velocity can be broken down into two parts: horizontal and vertical components. The horizontal component of the velocity, denoted as \(v_x\), is influenced by the initial speed and the angle of projection. Here, the angle given is \(\text{arctan}\frac{3}{4}\). In this specific case, using the trigonometric identity for cosine, we determine that \(\text{cos}(\theta) = \frac{4}{5}\). Multiplying this by the initial speed of \(20 \, \text{ms}^{-1}\), we get the horizontal component as \(16 \, \text{ms}^{-1}\).

Unlike the vertical component, the horizontal velocity remains constant throughout the projectile's flight if we neglect air resistance. So, even after 2 seconds, the horizontal component will stay at \(16 \, \text{ms}^{-1}\). This is a crucial point in understanding why the final answer to the problem is \(16 \, \text{ms}^{-1}\).

Gravity is a force that acts downward on any object, causing an acceleration of \(9.8 \, \text{ms}^{-2}\). It's important to note that gravity only influences the vertical component of the velocity. This means while the vertical speed changes over time due to gravity, the horizontal speed remains unaffected.

When calculating trajectories, always remember: gravity will affect the path of the projectile by accelerating it downwards, but it won't alter the rate at which it travels horizontally. In the context of our problem, since the initial horizontal velocity (\(16 \, \text{ms}^{-1}\)) is not influenced by gravity, it remains constant throughout the entire flight.

Breaking down a velocity into its trigonometric components is essential in projectile motion. Given an angle \( \theta \) to the horizontal, you can find the horizontal and vertical components of any initial velocity using sine and cosine functions:

• Horizontal component \(v_x = v \cos(\theta)\)
• Vertical component \(v_y = v \sin(\theta)\)

In our exercise, the angle is \(\text{arctan} \frac{3}{4}\). Here, we use the fact that \(\text{cos}(\theta) = \frac{4}{5}\) and \(\text{sin}(\theta) = \frac{3}{5}\), corresponding to a 3-4-5 right triangle. Therefore, the initial horizontal velocity component is calculated as \(16 \, \text{ms}^{-1}\) and the initial vertical component would be \(12 \, \text{ms}^{-1}\) if needed.

Understanding these components makes solving problems like this more manageable, as it clarifies how each directional force contributes to the overall motion.