Q 11-03-035NEETJEE MainMedium
A stone is projected at $30\sqrt{2}$ m/s at $45°$ to the horizontal. Its average velocity from the point of projection to the highest point is ($g = 10\ \text{m/s}^2$)
Answer: (D) $15\sqrt{5}$ m/s
Concept. Average velocity $= \dfrac{\text{displacement}}{\text{time}}$.
Step 1. Components. $u_x = u_y = 30$ m/s; time to the top $3$ s.
Step 2. Displacement. $x = 90$ m, $H = 45$ m: $\sqrt{90^2 + 45^2} = 45\sqrt5$ m.
Step 3. Average velocity. $\dfrac{45\sqrt5}{3} = 15\sqrt5$ m/s.
Answer. $15\sqrt5$ m/s (Option D).
Solution by Sreeraj P, M.Sc Physics