$A$ projectile is thrown with some initial velocity at an angle $\alpha$ to the horizontal. Its velocity when it is at the highest point is $(2/5)^{1/2}$ times the velocity when it is at height half of the maximum height. Find the angle of projection $\alpha$ with the horizontal. (in $^{\circ}$)

Two particles of equal masses are revolving in circular paths of radii $r_1$ and $r_2$ respectively with the same speed. The ratio of their centripetal forces is

$A$ 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).

$A$ ball of mass $100 \ g$ is projected with velocity $20 \ m/s$ at $60^{\circ}$ with the horizontal. The decrease in kinetic energy of the ball during the motion from the point of projection to the highest point is:

$A$ particle performs uniform circular motion with an angular momentum $L$. If the frequency of the particle's motion is doubled and its kinetic energy is halved, then its angular momentum becomes

$A$ boy throws a ball with a velocity $V_0$ at an angle $\alpha$ to the ground. At the same time,he starts running with a uniform velocity to catch the ball before it hits the ground. To achieve this,he should run with a velocity of