$A$ block of mass $5 \, kg$ is moving with a horizontal velocity of $1.5 \, m/s$. $A$ force of $5 \, N$ is applied in the perpendicular (vertical) direction for $4 \, s$. The total distance covered by the block is .......... $m$.

$A$ projectile is thrown upward at an angle $60^{\circ}$ with the horizontal. The speed of the projectile is $20 \text{ m/s}$ when its direction of motion is $45^{\circ}$ with the horizontal. The initial speed of the projectile is . . . . . . in $\text{m/s}$.

$A$ ball of mass $160 \, g$ is thrown up at an angle of $60^{\circ}$ to the horizontal at a speed of $10 \, m/s$. The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly $\left(g=10 \, m/s^{2}\right)$ (in $kg \cdot m^{2}/s$).

The angle of projection of a particle is measured from the vertical axis as $\phi$ and the maximum height reached by the particle is $h_m$. Here $h_m$ as a function of $\phi$ can be represented as:

$A$ 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 :

If a particle of mass $m$ covers half of the horizontal circle with constant speed $v$,then the change in its kinetic energy is