$A$ particle has an initial velocity of $(\hat{i} + \hat{j}) \, m/s$ and an acceleration of $(\hat{i} + \hat{j}) \, m/s^2$. Its speed after $10 \, s$ will be:

$A$ bullet fired at an angle of $30^{\circ}$ with the horizontal hits the ground $3.0 \; km$ away. By adjusting its angle of projection,can one hope to hit a target $5.0 \; km$ away? Assume the muzzle speed to be fixed,and neglect air resistance.

Two bodies are projected from the ground with the same speed $40 \ m/s$ at two different angles with respect to the horizontal. The bodies were found to have the same range. If one of the bodies was projected at an angle of $60^{\circ}$ with the horizontal,then the sum of the maximum heights attained by the two projectiles is $....... \ m$. (Given $g = 10 \ m/s^2$)

$A$ 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 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 :

$A$ particle has an initial velocity of $(3\hat i + 4\hat j) \; ms^{-1}$ and an acceleration of $(0.4\hat i + 0.3\hat j) \; ms^{-2}$. Its speed after $10 \; s$ is: