Question bank/JEE Main/Physics

At a height 0.4 m from ground, the velocity of projectile in
vector form
is$(\overset{\rightarrow}{v} = 6\overset{\hat{}}{i} + 2\overset{\hat{}}{j})$.
The angle of projection is π/n. Find n.

A6B3C2D4

Show answer and solution▾

${(v_{y})}^{2} = u_{y}^{2} + 2a_{y}S$
$(2)^{2} = {(u_{y})}^{2} + 2 \times ( - 10) \times 0.4$
$4 + 8 = u_{y}^{2}$
$u_{y} = \sqrt{12}$
$tan\theta = \frac{u_{y}}{u_{x}}$
$= \frac{\sqrt{12}}{6}$
$= \frac{1}{\sqrt{3}}$
$\theta = 30^{\circ}$

Source: Grass-G / P-18 Physics paper 27 Nov Third.docx