The velocity of projectile at the intial point $A$ is $\left( {2\hat i + 3\hat j} \right)$ $m/s $ . It's velocity (in $m/s$) at point $B$ is

Two objects are projected with same velocity ' $u$ ' however at different angles $\alpha$ and $\beta$ with the horizontal. If $\alpha+\beta=90^{\circ}$, the ratio of horizontal range of the first object to the $2^{\text {nd }}$ object will be :

Two particles are projected from the same point with the same speed at different angles $\theta _1$ and $\theta _2$ to the horizontal. They have the same range. Their times of flight are $t_1$ and $t_2$ respectively.

A projectile is fired at an angle of $45^o $ with the horizontal . Elevation angle of the projectile at its highest point as seen from the point of projection, is 

An arrow is shot into air. Its range is $200\,m$ and its time of flight is $5\,s$. If $g=10\,m / s ^2$, then horizontal component of velocity and the maximum height will be respectively

The maximum horizontal range of a projectile is $16\,km$ when the projectile is thrown at an elevation of $30^o$ from the horizontal, it will reach to the maximum height of ....... $km$