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.
${\theta _1} = 90 - {\theta _2}$
$\frac{{{t_1}}}{{\sin \,{\theta _1}}} = \frac{{{t_2}}}{{\sin \,{\theta _2}}}$
$\frac{{{t_1}}}{{{t_2}}} = \tan \,{\theta _1}$
All of the above
A stone projected with a velocity u at an angle $\theta$ with the horizontal reaches maximum height $H_1$. When it is projected with velocity u at an angle $\left( {\frac{\pi }{2} - \theta } \right)$ with the horizontal, it reaches maximum height $ H_2$. The relation between the horizontal range R of the projectile, $H_1$ and $H_2$ is
The speed of a projectile at its maximum height is $\frac {\sqrt 3}{2}$ times its initial speed. If the range of the projectile is $P$ times the maximum height attained by it, $P$ is equal to
In a projectile motion, velocity at maximum height is
The velocity at the maximum height of a projectile is half of its initial velocity $u$. Its range on the horizontal plane is
A particle $A$ is projected vertically upwards. Another identical particle $B$ is projected at an angle of $45^o $ . Both reach the same height. The ratio of the initial kinetic energy of $A$ to that of $B$ is