Q 11-03-031NEETJEE MainEasy
A ball is projected at $40$ m/s at $30°$ above the horizontal. Its maximum height and horizontal range are ($g = 10\ \text{m/s}^2$)
Answer: (D) $20$ m and $80\sqrt{3}$ m
Concept. Resolve the velocity; $H = \dfrac{u_y^2}{2g}$, $R = u_xT$.
Step 1. Components. $u_x = 20\sqrt3$ m/s, $u_y = 20$ m/s.
Step 2. Height and time. $H = 20$ m, $T = 4$ s.
Step 3. Range. $R = 20\sqrt3 \times 4 = 80\sqrt3$ m.
Answer. $20$ m and $80\sqrt3$ m (Option D).
Solution by Sreeraj P, M.Sc Physics