$A$ ball of mass $m$ is thrown vertically upwards. Another ball of mass $2m$ is thrown at an angle $\theta$ with the vertical. Both of them stay in air for the same period of time. The heights attained by the two balls are in the ratio of

The trajectory of a particle in projectile motion is given by $y = x - \frac{x^2}{80}$. Here, $x$ and $y$ are in meters. For this projectile motion, match the following with $g = 10 \, m/s^2$.

$Column-I$	$Column-II$
$(A)$ Angle of projection	$(p)$ $20 \, m$
$(B)$ Angle of velocity with horizontal after $4 \, s$	$(q)$ $80 \, m$
$(C)$ Maximum height	$(r)$ $45^{\circ}$
$(D)$ Horizontal range	$(s)$ $\tan^{-1}(1/2)$

The kinetic energy of a particle moving along a circle of radius $R$ depends on the distance $s$ as $K = as^2$,where $a$ is a constant. Then the force acting on the particle is

$A$ particle moves in the $X-Y$ plane under the influence of a force such that its linear momentum is $\vec{p}(t)=A[\hat{i} \cos (kt)-\hat{j} \sin (kt)]$,where $A$ and $k$ are constants. The angle between the force and the momentum is (in $^{\circ}$)

$A$ stone projected with a velocity $u$ at an angle $\theta$ with the horizontal reaches maximum height $H_1$. When it is projected with velocity $u$ at an angle $(\frac{\pi}{2} - \theta)$ with the horizontal,it reaches maximum height $H_2$. The relation between the horizontal range $R$ of the projectile,$H_1$ and $H_2$ is

$A$ particle has an initial velocity $(3\hat{i} + 4\hat{j})$ and an acceleration of $(0.4\hat{i} + 0.3\hat{j})$. Its speed after $10\,s$ is: