Projectile Motion

Per NCERT Class 11, Chapter 4 (Motion in a Plane), the maximum height of a projectile depends only on the vertical component of the initial velocity and is given by H = (u sinθ)² / (2g). The vertical component here is u sinθ = 20 × sin30° = 20 × 0.5 = 10 m/s. Substituting, H = (10)² / (2 × 10) = 100/20 = 5 m. The option 10 m incorrectly uses the full launch speed instead of its vertical component. The option 15 m has no valid derivation and overestimates the height. The option 2.5 m results from squaring 5 m/s, mistakenly halving the vertical velocity before substitution. Plausibility check: at maximum height the vertical velocity is zero, and the time to reach it is u sinθ/g = 1 s; the height covered in that time, (1/2)(10)(1)² subtracted appropriately, confirms 5 m, consistent with energy reasoning where vertical kinetic energy converts fully to potential energy.