91Ó°ÊÓ

When dealing with projectile motion, understanding initial velocity components is crucial as they dictate how the projectile moves both horizontally and vertically. In our scenario, the initial velocity is given as \((\hat{i} + 2 \hat{j}) \mathrm{m} / \mathrm{s}\). This expression breaks down into two parts: the horizontal component and the vertical component.
The horizontal component \(u_x\) is represented by the coefficient of \(\hat{i}\) which is 1 m/s, meaning the projectile moves at 1 m/s along the horizontal axis.
Similarly, the vertical component \(u_y\) is the coefficient of \(\hat{j}\), quantified as 2 m/s. This tells us the projectile initially moves up at a rate of 2 m/s.

These components are essential because they are used as inputs in later equations of motion to describe how the position of the projectile changes over time.

Equations of motion are mathematical formulas that describe the behavior of moving objects. In the context of projectile motion, they help us pinpoint the projectile's position over time in both horizontal and vertical planes.

For horizontal motion, the equation is relatively simple as there is no acceleration, since gravity acts only vertically. The equation is:

• \( x = u_x t \)

Which translates to \( x = t \) for our example, as \( u_x \) is 1 m/s. This means that horizontally, the object moves linearly with time.For vertical motion, gravity plays a role, and the equation is:

• \( y = u_y t - \frac{1}{2} g t^2 \)

Plugging the initial vertical velocity (2 m/s) and acceleration due to gravity (10 m/s²), we obtain: \( y = 2t - 5t^2 \) for the vertical movement. These equations indicate how the object's positions change separately in x and y directions over time before even considering combining them into a trajectory equation.

The trajectory equation combines both horizontal and vertical motions into a single relationship between coordinates on the projectile's path. Without involving time directly. This is done by eliminating the time variable from the position functions derived earlier.Initially, we have:

• For horizontal: \(x = t\)
  
• For vertical: \(y = 2t - 5t^2\)
  

To eliminate \(t\), we substitute \(t = x\) into the vertical motion equation to produce:

• \(y = 2x - 5x^2\)
  

This equation is the trajectory equation reflecting how height \(y\) changes with horizontal distance \(x\). It allows us to predict the projectile's path across a plane and is especially helpful in determining where it will land or any specific point it might reach.