(b) $Maximum\ horizontal\ range = 80\ m$ ∵ $\theta = 45{^\circ}\ m$ $\therefore\ \ \frac{u^{2}}{g} = 80\ m$ $Maximum\ height,\ h = \frac{u^{2}\sin^{2}\theta}{2g}$ $= \frac{80}{2}\left( \sin^{2}45{^\circ} \right) = 20\ m$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.