Projectile Motion Launched at an Angle | Height and Range | Grade 9 Science Quarter 4 Week 2

This lesson focuses on projectile motion launched at an angle, building on previous concepts. The objective is to investigate the relationship between the angle of release and the height and range of a projectile. Projectile motion combines constant horizontal velocity and vertical motion with constant acceleration (due to gravity).

Using a baseball example, the video explains that horizontal velocity (vx) remains constant, while vertical velocity (vy) changes. As the projectile rises (point A to B), vy decreases because gravity opposes the motion. At maximum height (point B), vy momentarily becomes zero. As it descends (point B to C), vy increases, aligning with gravity's direction.

Several facts are highlighted: objects are projected from rest at an upward angle, initial velocity resolves into horizontal and vertical components, horizontal velocity is constant (zero acceleration), the time to reach maximum height equals the time to return to launch height, and initial upward velocity magnitude equals final downward velocity magnitude at the same height.

The video demonstrates that a 45-degree angle results in the greatest horizontal range. A 75-degree angle achieves maximum height. Angles that are complementary (e.g., 30 and 60 degrees, or 15 and 75 degrees) result in the same range. As the launch angle increases, the vertical displacement also increases. At the highest point, the vertical component of velocity is zero, and the time to reach this point is half of the total flight time.

An example problem is presented: a baseball hit at 25 degrees with a velocity of 30 m/s. The first step is to find the maximum height reached. Using the formula (vi * sin(theta))^2 / (2 * g), with vi = 30 m/s, theta = 25 degrees, and g = 9.8 m/s^2, the calculation shows the maximum height is 8.20 meters.

The second part of the example problem involves finding the horizontal displacement (range). First, the total time of flight needs to be calculated using the formula 2 * vi * sin(theta) / g. This yields a total time of 2.59 seconds. Then, the horizontal displacement (dx) is calculated by multiplying vi * cos(theta) * time, resulting in a range of 70.42 meters.