Range of a bullet fired at $45^o$ to the horizontal is $980 \, m$. If the bullet is fired at the same angle from a car travelling horizontally at $18 \, km/h$ towards the target,then the range will be increased by:

$A$ car is going round a circle of radius $R_1$ with constant speed. Another car is going round a circle of radius $R_2$ with constant speed. If both of them take the same time to complete the circles,the ratio of their angular speeds and linear speeds will be .........

$A$ particle starts from the origin at $t=0$ with a velocity $5 \hat{i} \text{ m/s}$ and moves in the $x-y$ plane under the action of a force which produces a constant acceleration of $(3 \hat{i} + 2 \hat{j}) \text{ m/s}^2$. If the $x$-coordinate of the particle at that instant is $84 \text{ m}$,then the speed of the particle at this time is $\sqrt{\alpha} \text{ m/s}$. The value of $\alpha$ is . . . . . . .

Two projectiles are thrown with the same range $R$. If their maximum heights are $h_1$ and $h_2$ respectively,then what is the value of the range $R$?

$A$ particle of mass $m$ performs uniform circular motion of radius $r$ with linear speed $v$ under the application of force $F$. If $m$,$v$,and $r$ are all increased by $20 \%$,the necessary change in force required to maintain the particle in uniform circular motion is: (in $\%$)

$A$ particle of mass $m$ is moving in the $xy$-plane such that its velocity at a point $(x, y)$ is given as $\vec{v} = \alpha(y \hat{i} + 2x \hat{j})$,where $\alpha$ is a non-zero constant. What is the force $\vec{F}$ acting on the particle?