The trajectory of a projectile in a vertical plane is y = $$\alpha$$x $$-$$ $$\beta$$x², where $$\alpha$$ and $$\beta$$ are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection $$\theta$$ and the maximum height attained H are respectively given by :

When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed v, he sees that rain drops are coming at an angle 60° from the horizontal. On further increasing the speed of the car to (1 + $$\beta $$)v, this angle changes to 45^o. The value of $$\beta $$ is close to :

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: