← Mechanics
Kinematics & projectiles (every paper)
Signs first. Horizontal constant, vertical g = 9.8 down.Core equations
No time v² = u² + 2as Velocity–time v = u + at Displacement s = ut + ½at² Projectile range x = v·t (horizontal) Exam-style Ball kicked at 12 m s⁻¹ at 30° above horizontal Find time in the air and range (g = 9.8). Vertical: u_y = 12 sin30° = 6.0 m s⁻¹. At top v_y = 0 → t_up = 6.0/9.8 = 0.61 s. Total time = 2 × 0.61 = 1.22 s. Horizontal: v_x = 12 cos30° = 10.4 m s⁻¹; range = 10.4 × 1.22 = 12.7 m. ✓ t ≈ 1.2 s, range ≈ 13 m. State the split into components explicitly.Reading a "without calculating, compare" projectile question
- Time in the air depends ONLY on the vertical launch speed and g — mass never appears in the constant-acceleration equations, so a heavier object launched identically takes exactly the same time.
- Horizontal range depends on both the horizontal speed AND the time in the air — if either changes, range changes with it.
- A steeper launch angle (closer to 90°) increases time in the air but can decrease range, since range trades off between vertical time and horizontal speed as the angle changes.