91Ó°ÊÓ

The launch angle plays a crucial role in projectile motion. It is the angle at which a projectile is launched relative to the horizontal ground. This angle affects both the range and the height of the projectile.

When the angle is large, the projectile follows a steeper path, reaching a higher peak before descending. Conversely, a smaller launch angle results in a flatter trajectory with a longer horizontal distance but a lower maximum height.

In this exercise, the launch angle of the projectile was set at \(15.0^{\circ}\). This angle is relatively shallow, meaning the projectile will travel further horizontally rather than vertically, but must still clear the barrier height of 13.5 meters.

Initial velocity is the speed at which a projectile is launched. It has two components: horizontal and vertical. Both components are affected by the launch angle.

The horizontal component \( v_{x0} \) can be calculated as \( v_0 \cos(\theta) \), while the vertical component \( v_{y0} \) is \( v_0 \sin(\theta) \). Here, \( v_0 \) is the overall initial launch speed, and \( \theta \) is the launch angle.

In our example, we calculated an initial velocity \( v_0 \) using the vertical motion equation. It was determined that the projectile needs a speed of approximately 43.0 m/s to clear the 13.5 m barrier when launched at a \(15^{\circ}\) angle.

The vertical motion of a projectile is governed by the laws of physics and can be described with a specific equation. This equation helps find unknown variables like initial velocity or maximum height.

The key vertical motion equation used in this problem is:\[ v_{y}^2 = v_{y0}^2 - 2gh \]where:

• \( v_{y} \) is the final vertical velocity at maximum height, which is zero.
• \( v_{y0} \) is the initial vertical velocity, calculated as \( v_0 \sin(\theta) \).
• \( g \) is the acceleration due to gravity (approximately 9.8 m/s²).
• \( h \) is the maximum height the projectile reaches.

In the original step-by-step solution, this equation was rearranged and solved to find the initial speed \( v_0 \) required to reach the specified height of 13.5 meters.