$A$ particle moves with constant angular velocity in a circle. During the motion,its:

Two balls of same mass are projected with the same speed, one vertically upwards and the other at an angle of $60^{\circ}$ with the vertical. The ratio of their potential energies at the highest point is:

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

Two bodies were thrown simultaneously from the origin: one straight up and the other at an angle of $60^{\circ}$ to the vertical. The initial velocity of each body is $10 \ m \ s^{-1}$. Neglecting air resistance, the distance between the two bodies after $t=2 \ s$ is (Use $g=10 \ m \ s^{-2}$):

$A$ body of mass $100 \, g$ is projected at an angle of $30^\circ$ with the horizontal with a velocity of $20 \, m \, s^{-1}$. What is the change in its momentum at the maximum height in $kg \, m \, s^{-1}$?

$A$ 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 behaviors for launch angles $\alpha_1$ and $\alpha_2$ as shown in the figure. 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 a critical angle $\alpha_c$ (where $\alpha_1 < \alpha_c < \alpha_2$). The value of $\alpha_c$ is (where $v_0$ is the initial speed of the projectile and $g$ is the acceleration due to gravity).