At any instant $t$, the vertical distance $Y$ and horizontal distance $X$ of a projectile are given by $2Y = 6t - gt^2$ and $X = 4t$. The initial velocity of the projectile is ($X$ and $Y$ are in $m$ and $t$ is in $s$). (in $\,m/s$)

$A$ projectile is thrown with a velocity of $50 \, ms^{-1}$ at an angle of $53^o$ with the horizontal. The equation of the trajectory is given by:

$A$ gun and a target are at the same horizontal level separated by a distance of $600 \,m$. The bullet is fired from the gun with a velocity of $500 \,ms^{-1}$. In order to hit the target, the gun should be aimed to a height $h$ above the target. The value of $h$ is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $\,m$)

$A$ body of mass $m$ is thrown upwards at an angle $\theta$ with the horizontal with velocity $v$. While rising up,the velocity of the mass after $t$ seconds will be:

$A$ 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].

$A$ particle is thrown with a velocity of $u \, m/s$ at an angle of $30^{\circ}$ with the horizontal. It passes through points $A$ and $B$ at the same vertical height at times $t_1 = 1 \, s$ and $t_2 = 3 \, s$ respectively,as shown in the figure. The value of $u$ is ....... $m/s$ (take $g = 10 \, m/s^2$).