A body is projected with velocity $u$ making an angle $\alpha$ with the horizontal. Its velocity when it is perpendicular to the initial velocity vector $u$ is

A body is projected at an angle $\theta $ with horizontal, another body is projected with the same speed at an angle $\theta $ with the vertical then the ratio of the maximum height is

The maximum horizontal range of a projectile is $160\, m$. When the projectile is thrown  with the same speed at an elevation of $30^o$ from the horizontal, it will reach to the  maximum height of   ......... $m$

At the highest point of the path of a projectile, its

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 particle is projected from the ground with velocity $u$ at angle $\theta$ with horizontal. The horizontal range, maximum height and time of flight are $R, H$ and $T$ respectively. They are given by $R = \frac{{{u^2}\sin 2\theta }}{g}$, $H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ and $T = \frac{{2u\sin \theta }}{g}$  Now keeping $u $ as fixed, $\theta$ is varied from $30^o$ to $60^o$. Then,