Advanced Projectile Motion Analysis

This page delves deeper into projectile motion for A-Level Physics, focusing on a projectile launched at a 45° angle with an initial velocity of 45 m/s.

The analysis begins by resolving the initial velocity into its horizontal and vertical components using trigonometry:

• Vertical component: 45 sin(27°)
• Horizontal component: 45 cos(27°)

Vocabulary: Resolving vectors means breaking them down into their horizontal and vertical components.

The page then applies SUVAT equations to calculate key parameters:

1. Maximum height: Using v^2 = u^2 + 2as with v = 0 at the peak
2. Time to reach maximum height: Using v = u + at
3. Total time of flight: Doubling the time to reach maximum height due to symmetry
4. Horizontal range: Using s = ut for the horizontal motion

Highlight: The trajectory is symmetrical, so the time to reach maximum height is half the total flight time.

The analysis concludes by noting that the angle of impact with the ground is equal to the launch angle due to the symmetry of the parabolic path.

Example: For a projectile launched at 45 m/s at 27°, the maximum height is 21.3 m, the time of flight is 4.17 s, and the range is 167 m.

This page reinforces the importance of vector resolution and symmetry in solving A-Level Physics projectile motion questions.