$A$ stone is projected from the ground at $t = 0$. At the time of projection,the horizontal and vertical components of velocity are $10\, m/s$ and $20\, m/s$ respectively. Find the time at which the magnitudes of tangential and normal acceleration will be equal $(g = 10\, m/s^2)$ [neglect air friction].

$A$ body projected at a certain angle $(\neq 90^{\circ})$ from the ground crosses a point in its path at a time $t_1 = 2.3 \ s$ and from there it reaches the ground after a further time $t_2 = 5.7 \ s$. The maximum height reached by the body is (Take $g = 10 \ ms^{-2}$) (in $m$)

In a projectile motion,the velocity at maximum height is:

$A$ stone is projected vertically upwards with speed '$v$'. Another stone of same mass is projected at an angle of $60^{\circ}$ with the vertical with the same speed '$v$'. The ratio of their potential energies at the highest points of their journey is $\left[\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]$

Average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory is: (projection speed = $u$,angle of projection from horizontal = $\theta$)

$A$ particle $P$ is projected with velocity $u_1$ at an angle of $30^\circ$ with the horizontal. Another particle $Q$ is projected vertically upwards with velocity $u_2$ from the point directly below the maximum height of $P$. The condition for the collision of both particles is: