If T is the total time of flight, $h$ is the maximum height $ \& R$ is the range for horizontal motion, the $x$ and $y$ co-ordinates of projectile motion and time $t$ are related as
$y = 4h\left( {\frac{t}{T}} \right)\,\,\,\left( {1\,\, - \,\,\frac{t}{T}} \right)$
$y = 4h\left( {\frac{X}{R}} \right)\,\,\,\left( {1\,\, - \,\,\frac{X}{R}} \right)$
$y = 4h\left( {\frac{T}{t}} \right)\,\,\,\left( {1\,\, - \,\,\frac{T}{t}} \right)$
Both $(A)$ and $(B)$
A projectile cover double range as compare to its maximum height attained. The angle of projection is
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 particle is projected from the ground at an angle of $\theta $ with the horizontal with an initial speed of $u$. Time after which velocity vector of the projectile is perpendicular to the initial velocity is
Galileo writes that for angles of projection of a projectile at angles $(45 + \theta )$ and $(45 - \theta )$, the horizontal ranges described by the projectile are in the ratio of (if $\theta \le 45)$
A particle is projected vertically upwards from $O$ with velocity $v$ and a second particle is projected at the same instant from $P$ (at a height h above $O$) with velocity $v$ at an angle of projection $\theta$ . The time when the distance between them is minimum is