$A$ particle moves over a $xy$ plane with a constant acceleration $\vec{a} = (4.0 \, m \, s^{-2}) \hat{i} + (4.0 \, m \, s^{-2}) \hat{j}$. At time $t = 0$,the velocity is $\vec{v}_0 = (4.0 \, m \, s^{-1}) \hat{i}$. The speed of the particle when it is displaced by $6.0 \, m$ parallel to the $x$-axis is,

$A$ particle of mass $m$ is projected at a velocity $u$,making an angle $\theta$ with the horizontal ($x$-axis). If the angle of projection $\theta$ is varied keeping all other parameters same,then the magnitude of angular momentum $(L)$ at its maximum height about the point of projection varies with $\theta$ as,

$A$ projectile is thrown upward with a velocity $v_0$ at an angle $\alpha$ to the horizontal. The change in velocity of the projectile when it strikes the same horizontal plane is

Two particles,one at the centre of a circle of radius $R$,and another at a point $Q$ on the circle,start moving towards a point $P$ on the circle at the same time (see figure below). Both move with uniform velocities $\vec{V}_1$ and $\vec{V}_2$ respectively. They reach the point $P$ at the same time. If the angle between the velocities is $\theta$ and the angle subtended by $P$ and $Q$ at the centre is $\phi$ (as shown in the figure),then:

$A$ particle crossing the origin at time $t=0$,moves in the $xy$-plane with a constant acceleration $a$ in the $y$-direction. If the equation of motion of the particle is $y = bx^2$ (where $b$ is a constant),then its velocity component in the $x$-direction is

Two balls are thrown simultaneously from the ground with the same velocity of $10\,m/s$ but at different angles of projection with the horizontal. Both balls fall at the same distance $5\sqrt{3}\,m$ from the point of projection. What is the time interval between the balls striking the ground?