A projectile is launched from ground level with an initial speed of 20 m/s at 30° above horizontal. What is the maximum height reached? Use g = 9.8 m/s² and ignore air resistance.
A2.55 m
B5.10 m
C15.3 m
D20.4 m
Explanation
📌 Vertical component v_iy = v sin θ = 20 × sin(30°) = 10 m/s. At max height v_y = 0. Use v_fy² − v_iy² = −2g·h → 0 − 100 = −19.6h → h = 5.10 m. Distractor A used v_iy/2 (halved velocity); C used cos(30°) instead of sin(30°) — the classic sin/cos swap; D ignored angle entirely, used full v² / 2g.