91Ó°ÊÓ

The initial angle of projection refers to the angle at which an object, like a projectile, is launched with respect to the horizontal. This angle is crucial because it determines the path or trajectory that the object will follow. In our exercise, at \( t=2 \) seconds, we have information about the projectile's speed and the angle it makes with the horizontal, but we need to find the angle of projection or the initial launch angle. To do this, we need to consider both the vertical and horizontal components of the velocity at that moment.

When a projectile is launched, it has an initial velocity that can be broken into two components:

• Horizontal Component (v_x): This remains constant throughout the motion, assuming no air resistance.
• Vertical Component (v_y): This is affected by gravity, causing the projectile to eventually fall back to the ground.

Understanding these components helps in calculating the initial angle as they change over time due to gravity's influence.

In projectile motion, the overall velocity of an object can be split into two components: horizontal and vertical. At \( t=2 \) seconds, we are told the projectile's speed is \( 20 \ ext{ m/s} \) and it's moving at \( 30^{\circ} \) relative to the horizontal. From these details, we can deduce the velocity components:

The **Horizontal Velocity Component** is given by:\[ v_x = v \cos(30^{\circ}) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \text{ m/s} \]This component is constant if we ignore air resistance. On the other side, the **Vertical Velocity Component** is determined by:\[ v_y = v \sin(30^{\circ}) = 20 \times \frac{1}{2} = 10 \text{ m/s} \]The vertical component is influenced by gravity, and at the \( t=0 \) time of projection, it might be different. In our case, gravity affects the velocity as:\[ v_y = u\sin(\theta) - gt \]Therefore, understanding these two components is essential for solving the initial angle of projection, as the relation between these components at time \( t=2 \) gives insights into the initial conditions.

Trigonometry plays a vital role in analyzing projectile motion, as it helps relate the components of velocity to the initial angle of projection. Since projectiles move in two dimensions, trigonometric functions like sine, cosine, and tangent help describe this motion.

Using Sine and Cosine

In our exercise, the sine and cosine of the angle give the vertical and horizontal velocity components, respectively:

• The **sine** function: \( \sin(\theta) = \frac{\text{opp}}{\text{hyp}} \) helps find the component of velocity in the vertical direction, which is the opposite side.
• The **cosine** function: \( \cos(\theta) = \frac{\text{adj}}{\text{hyp}} \) relates to the horizontal or adjacent side velocity component.

Tangent and Angle Calculation

The tangent function is particularly helpful for explicitly finding the angle of projection. By using:\[ \tan(\theta) = \frac{u\sin(\theta)}{u\cos(\theta)} \]We can solve for the initial angle. In this exercise, we find:\[ \tan(\theta) = \frac{30}{10\sqrt{3}} = \sqrt{3} \]Thus, resulting in:\[ \theta = \tan^{-1}(\sqrt{3}) = 60^{\circ} \]Using these trigonometric tools, students can effectively analyze the components of velocity and their relationship to the angle of projection.