$A$ body is moving with constant speed in a circle of radius $10 \ m$. The body completes one revolution in $4 \ s$. At the end of the $3^{rd}$ second,the displacement of the body (in $m$) from its starting point 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:

$A$ ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal surface. It reaches a maximum height of $120 \ m$ and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of $30^{\circ}$ with the horizontal surface. The maximum height it reaches after the bounce, in metres, is.

$A$ point moves in the $xy$-plane according to the following equations: $x = a \sin \omega t$ and $y = a(1 - \cos \omega t)$, where $a$ and $\omega$ are positive constants. Find the angle between the point's velocity and acceleration vectors.

$A$ hill is $500 \, m$ high. Supplies are to be sent across the hill,using a cannon that can hurl packets at a speed of $125 \, m/s$ over the hill. The cannon is located at a distance of $800 \, m$ from the foot of the hill and can be moved on the ground at a speed of $2 \, m/s$; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach the ground on the other side of the hill? Take $g = 10 \, m/s^2$.

$A$ 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 :