A particle moves from the point $\left( {2.0\hat i + 4.0\hat j} \right)\,m$, at $t = 0$ with an initial velocity $\left( {5.0\hat i + 4.0\hat j} \right)\,m{s^{ - 1  }}$. It is acted upon by a constant force which produces a constant acceleration $\left( {4.0\hat i + 4.0\hat j} \right)\,m{s^{ - 2}}$. What is the distance of the particle from the origin at time $2\,s$

$Assertion$ : If a body is thrown upwards, the distance covered by it in the last second of upward motion is about $5\, m$ irrespective of its initial speed
$Reason$ : The distance covered in the last second of upward motion is equal to that covered in the first second of downward motion when the particle is dropped.

Particles $A$ and $B$ are moving with constant velocities along $x$ and $y$ axis respectively, the graph of separation between them with time is

A projectile is projected with speed $u$ at an angle $\theta$ with the horizontal. The average velocity of the projectile between the instants it crosses the same level is ............

A projectile is fired from horizontal ground with speed $v$ and projection angle $\theta$. When the acceleration due to gravity is $g$, the range of the projectile is $d$. If at the highest point in its trajectory, the projectile enters a different region where the effective acceleration due to gravity is $g^{\prime}=\frac{g}{0.81}$, then the new range is $d^{\prime}=n d$. The value of $n$ is. . . . .

The figure shows a velocity-time graph of a particle moving along a straight line  If the particle starts from the position $x_0=-15\,m$ , then its position at $t=2\,s$ will be   ........ $m$