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

$A$ projectile is launched with an initial speed of $40 \ m \ s^{-1}$ at an angle of $30^{\circ}$ above the ground. The projectile lands on a hillside $2.0 \ s$ later. The net displacement from where the projectile was launched to where it hits the target is (Take $g = 10 \ m \ s^{-2}$)

$A$ particle at a distance of $1 \ m$ from the origin starts moving,such that $dr/d\theta = r$,where $r$ and $\theta$ are polar coordinates. Then,the angle between the resultant velocity and the tangential velocity is

$A$ car is moving with a velocity of $4 \,m \,s^{-1}$ towards east. After a time of $4 \,s$, if it is heading north-east with a velocity of $4 \sqrt{2} \,m \,s^{-1}$, then the average velocity of the car is

The figure shows the velocity and the acceleration of a point-like body at the initial moment of its motion. The direction and the absolute value of the acceleration remain constant. Find the time when the speed becomes minimum. (Given: $a = 4\, m/s^2$,$v_0 = 40\, m/s$,$\phi = 143^o$)

$A$ point moves in the $x-y$ plane according to the law $x = 3 \cos 4t$ and $y = 3(1 - \sin 4t)$. The distance travelled by the particle in $2 \text{ s}$ is...........$m$ (where $x$ and $y$ are in meters).