Neglecting the air resistance, the time of flight of a projectile is determined by
${U_{vertical}}$
${U_{horizontal}}$
$U = {U^2}_{vertical} + {U^2}_{horizontal}$
$U = U\,{({U^2}_{vertical} + {U^2}_{horizontal})^{1/2}}$
A ball is projected vertically upwards with a certain initial speed. Another ball of the same mass is projected with the same speed at an angle of $30^o$ with the horizontal. At the highest point, the ratio of their potential energies is
A projectile is thrown with a velocity of $50\,\, ms^{^{-1}}$ at an angle of $53^o$ with the horizontal Determine the instants at which the projectile is at the same height
At the top of the trajectory of a projectile, the directions of its velocity and acceleration are
A projectile crosses two walls of equal height $H$ symmetrically as shown The velocity of projection is........ $ms^{-1}$
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]