Maximum height in projectile motion

Steps:

• Initial vertical velocity: u_y = 12 sin(50°) ≈ 12 × 0.766 = 9.19 m/s
• At maximum height, vertical velocity v_y = 0
• Use equation: v_y² = u_y² - 2gh → h = u_y² / (2g) with g = 9.8 m/s²
• h = (9.19)² / (2 × 9.8) ≈ 84.5 / 19.6 ≈ 4.3 m

Why C is correct:

• C equals h = (v sin θ)² / (2g), the standard formula for maximum height from vertical motion kinematics.

Why the others are wrong:

• A: Likely uses horizontal component or cos(50°) instead of sin(50°).
• B: Underestimates by using g = 10 m/s² and rounded sin(50°) ≈ 0.7.
• D: Overestimates, possibly from total range formula or v² / (2g) without angle.

Final answer: C