$A$ particle is moving in $x-y$ plane according to $\vec{r} = b \cos \omega t \hat{i} + b \sin \omega t \hat{j}$,where $\omega$ is a constant. Which of the following statement$(s)$ is/are true?

$A$ particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that:

The angle of projection of a particle is measured from the vertical axis as $\phi$ and the maximum height reached by the particle is $h_m$. Here $h_m$ as a function of $\phi$ can be represented as:

$A$ particle moves along a circle of radius $R$ with kinetic energy $K = as^2$,where $s$ is the distance traveled. What is the net force acting on the particle?

$A$ 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).

$A$ scooter going due east at $10\, ms^{-1}$ turns right through an angle of $90^\circ$. If the speed of the scooter remains unchanged in taking the turn,the change in the velocity of the scooter is