A $NCC$ parade is going at a uniform speed of $9\,km / h$ under a mango tree on which a monkey is sitting at a height of $19.6\,m$. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is $...m$

(Given $g=9.8\,m / s ^{2}$ )

A particle has initial velocity $(2\hat i + 3\hat j ) $ and has acceleration $(0.3\,\hat i + 0.2\,\hat j)$ . Its speed after $10\,s$ is

Which physical quantity can be found by first differntiation and second differentiation of position vector ?

Let $\vec v$ and $\vec a$ denote the velocity and acceleration respectively of a body in one-dimensional motion 

Ship $A$ is sailing towards north -east with velocity $\vec v = 30\,\hat i + 50\hat j\,km/hr$ where $\hat i$ points east and $\hat j$ , north. Ship $B$ is at a distance of $80\, km$ east and $150\, km$ north of Ship $A$ and is sailing towards west at $10\, km/hr$. $A$ will be at minimum distance from $B$ is.........$hrs$

For any arbitrary motion in space, which of the following relations are true

$(a)$ $\left. v _{\text {average }}=(1 / 2) \text { (v }\left(t_{1}\right)+ v \left(t_{2}\right)\right)$

$(b)$ $v _{\text {average }}=\left[ r \left(t_{2}\right)- r \left(t_{1}\right)\right] /\left(t_{2}-t_{1}\right)$

$(c)$ $v (t)= v (0)+ a t$

$(d)$ $r (t)= r (0)+ v (0) t+(1 / 2)$ a $t^{2}$

$(e)$ $a _{\text {merage }}=\left[ v \left(t_{2}\right)- v \left(t_{1}\right)\right] /\left(t_{2}-t_{1}\right)$

(The 'average' stands for average of the quantity over the time interval $t_{1}$ to $t_{2}$ )