$A$ man standing on the roof of a house of height $h$ throws one particle vertically downwards and another particle horizontally with the same velocity $u$. The ratio of their velocities when they reach the earth's surface will be

$A$ 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).

$A$ particle moves with a constant speed of $5 \, ms^{-1}$ as shown in the figure. What is the change in velocity during half a revolution in $ms^{-1}$?

$A$ particle moving in a circle of radius $R$ with uniform speed takes time $T$ to complete one revolution. If this particle is projected with the same speed at an angle $\theta$ to the horizontal,the maximum height attained by it is equal to $4 R$. The angle of projection $\theta$ is then given by :

Two particles move on a circular path (one just inside and the other just outside) with angular velocities $\omega$ and $5\omega$ starting from the same point. Then

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