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
$u_x = u_y = 30$ m/s. Time to the top: $t = \dfrac{30}{10} = 3$ s.
At the top: $x = 30 \times 3 = 90$ m, $H = \dfrac{30^2}{20} = 45$ m.
$$|\text{displacement}| = \sqrt{90^2 + 45^2} = 45\sqrt{5}\ \text{m}, \qquad v_{avg} = \frac{45\sqrt{5}}{3} = 15\sqrt{5}\ \text{m/s}$$
Solution by Sreeraj P, M.Sc Physics