91Ó°ÊÓ

When a projectile is thrown in an upward direction, the angle it makes with the horizontal line is known as the inclination angle. This angle is crucial as it affects the projectile's trajectory and distance it will cover. In our exercise, the initially given inclination is \(60^{\circ}\), representing the angle at which the projectile is launched. As the projectile moves, this angle changes depending on the velocities in horizontal and vertical directions. A critical part of understanding projectile motion is recognizing that the inclination angle impacts how the velocities in these directions interact.

For example, the problem asks when the projectile's inclination will be \(45^{\circ}\). Reaching this angle signifies a specific balance between downward gravity influence and forward motion. At \(45^{\circ}\), the vertical and horizontal velocity components are equal. Understanding this concept helps in predicting the projectile's behavior throughout the motion.

When a projectile is thrown, its velocity can be broken down into two components: horizontal and vertical. These components originate from the initial velocity paired with the inclination angle.

• **Horizontal component** \((v_{0x})\): Calculated as \(v_0 \cos(\text{angle})\). In our situation, the angle is \(60^{\circ}\), so \(v_{0x} = 147 \cdot \frac{1}{2} = 73.5\ \mathrm{m/s}\). This component remains constant as the projectile moves because no forces are acting in the horizontal direction under ideal conditions.
• **Vertical component** \((v_{0y})\): Obtained using \(v_0 \sin(\text{angle})\). Here, \(v_{0y} = 147 \cdot \frac{\sqrt{3}}{2} = 73.5\sqrt{3}\,\mathrm{m/s}\). Unlike the horizontal component, the vertical component varies with time due to gravitational acceleration (\(g = 9.8\ \mathrm{m/s^2}\)).

These components form the foundation for analyzing projectile motion, enabling us to express the projectile's position and velocity at any given moment.

Time of flight refers to the duration a projectile spends in the air from the moment of launch until it lands. While this particular exercise focuses on the time it takes for the projectile to achieve a specific inclination, understanding time of flight in general aids in comprehending motion dynamics.

In the problem, finding the time when the inclination angle becomes \(45^{\circ}\) involves setting the tangent of the inclination equal to 1, as \(\tan(45^{\circ}) = 1\). The equation \(\frac{v_y}{v_x} = 1\) is used to determine when vertical velocity equals the horizontal one. By substituting the expressions for \(v_y\) and \(v_x\), we derive:

\[73.5\sqrt{3} - 9.8t = 73.5\]

This leads to:

\[t = \frac{73.5(\sqrt{3} - 1)}{9.8}\]

After simplifying:

\[t = 7.5(\sqrt{3} - 1)\, \mathrm{s}\]

This calculation shows how time of flight relates closely to the interplay between velocity components and gravity, essential for predicting motion outcomes.