When a force of constant magnitude and a fixed direction acts on a moving object,then its path is

Four persons $K, L, M$ and $N$ are initially at the corners of a square of side length $d$. If every person starts moving with speed $v$ such that $K$ is always headed towards $L$,$L$ towards $M$,$M$ towards $N$,and $N$ towards $K$,then the four persons will meet after

$A$ projectile is thrown at an angle $\theta$ to the horizontal from $O$ and it hits the ground at $O'$. During the entire motion (select the wrong statement):

$A$ 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:

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 $g=10 \ m \ s^{-2}$):

$A$ projectile is fired from the surface of the earth with a velocity of $5 \, m s^{-1}$ and angle $\theta$ with the horizontal. Another projectile is fired from another planet with a velocity of $3 \, m s^{-1}$ at the same angle and follows a trajectory which is identical to the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in $m s^{-2}$) (Given $g = 9.8 \, m s^{-2}$):