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
$u_x = 40\cos 30° = 20\sqrt{3}$ m/s, $u_y = 40\sin 30° = 20$ m/s.
$$H = \frac{u_y^2}{2g} = \frac{400}{20} = 20\ \text{m}, \qquad T = \frac{2u_y}{g} = 4\ \text{s}$$
$$R = u_xT = 20\sqrt{3} \times 4 = 80\sqrt{3}\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics