A projectile is thrown at an angle $\theta$ such that it is just able to cross a vertical wall at its highest point as shown in the figure.The angle $\theta$ at which the projectile is thrown is given by

A projectile is launched from the origin in the $xy$ plane ( $x$ is the horizontal and $y$ is the vertically up direction) making an angle $\alpha$ from the $x$-axis. If its distance. $r =\sqrt{ x ^2+ y ^2}$ from the origin is plotted against $x$, the resulting curves show different behaviours for launch angles $\alpha_1$ and $\alpha_2$ as shown in the figure below. For $\alpha_1, r ( x )$ keeps increasing with $x$ while for $\alpha_2$, $r(x)$ increases and reaches a maximum, then decreases and goes through a minimum before increasing again. The switch between these two cases takes place at an angle $\alpha_c\left(\alpha_1 < \alpha_c < \alpha_2\right)$. The value of $\alpha_c$ is [ignore where $v_0$ is the initial speed of the projectile and $g$ is the acceleration due to gravity]

Three identical balls are projected with the same speed at angle $30^o, 45^o$ and $60^o$. Their ranges are $R_1 R_2$ and $R_3$ respectively. Then

A projectile has the same range $R$ for two angles of projection. If $T_1$ and $T_2$ be the times of flight in the two cases, then $R$ is

Derive the formula for Range of a projectile $(R)$. Derive the formula for maximum projectile.