A balloon is moving up in air vertically above a point A on the ground. When it is at a height h₁, a girl standing at a distanced (point B) from A (see figure) sees it at an angle 45^o with respect to the vertical. When the balloon climbs up a further height h₂, it is seen at an angle 60^o with respect to the vertical if the girl moves further by a distance 2.464 d(point C). Then the height h₂ is (given tan 30^o = 0.5774)

Starting from the origin at time t = 0, with initial velocity 5$$\widehat j$$ ms^-1 , a particle moves in the x-y plane with a constant acceleration of $$\left( {10\widehat i + 4\widehat j} \right)$$ ms^-2. At time t, its coordinates are (20 m, y₀ m). The values of t and y₀ are, respectively:

A particle starts from the origin at t = 0 with an
initial velocity of 3.0 $$\widehat i$$ m/s and moves in the
x-y plane with a constant acceleration $$\left( {6\widehat i + 4\widehat j} \right)$$ m/s² . The x-coordinate of the particle at the instant when its y-coordinate is 32 m is D meters. The value of D is :-

A particle moves such that its position vector $$\overrightarrow r \left( t \right) = \cos \omega t\widehat i + \sin \omega t\widehat j$$ where $$\omega $$ is a constant and t is time. Then which of the following statements is true for the velocity $$\overrightarrow v \left( t \right)$$ and acceleration $$\overrightarrow a \left( t \right)$$ of the particle :