$A$ projectile is thrown into space so as to have a maximum possible horizontal range of $400 \, m$. Taking the point of projection as the origin,the coordinates of the point where the velocity of the projectile is minimum are

The position of a projectile launched from the origin at $t = 0$ is given by $\vec{r} = (40\hat{i} + 50\hat{j})\,m$ at $t = 2\,s$. If the projectile was launched at an angle $\theta$ from the horizontal,then $\theta$ is (take $g = 10\,m/s^2$)

It is possible to project a particle with a given velocity in two possible ways so as to make them pass through a point $P$ at a horizontal distance $r$ from the point of projection. If $t_1$ and $t_2$ are times taken to reach this point in two possible ways,then the product $t_1 t_2$ is proportional to

If the time of flight of a bullet over a horizontal range $R$ is $T$,then the angle of projection with the horizontal is ......

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: When a body is projected at an angle $45^{\circ}$,its range is maximum.
Reason $R$: For maximum range,the value of $\sin 2\theta$ should be equal to one.
In the light of the above statements,choose the correct answer from the options given below:

The initial velocity of a particle of mass $2\,kg$ is $(4 \hat{i} + 4 \hat{j})\,m/s$. $A$ constant force of $-20 \hat{j}\,N$ is applied on the particle. Initially,the particle was at $(0,0)$. Find the $x$-coordinate of the point where its $y$-coordinate is again zero. $..........\,m$