Two projectiles of same mass and with same velocity are thrown at an angle $60^o$ and $30^o$ with the horizontal,then which quantity will remain same?

$A$ batsman hits a ball with a velocity '$v$',making an angle of $60^{\circ}$ with the vertical. After some time, the direction of velocity makes an angle of $60^{\circ}$ with the horizontal. The speed of the ball at this instant is $[\cos(60^{\circ}) = \frac{1}{2}, \cos(30^{\circ}) = \frac{\sqrt{3}}{2}]$.

The equation for the trajectory of a projectile is $y = \left( \frac{x}{\sqrt{3}} - \frac{x^2}{60} \right) \text{ m}$. The velocity of projection of the projectile is (Acceleration due to gravity $g = 10 \text{ m s}^{-2}$) (in $\text{ m s}^{-1}$)

$A$ body is projected from the earth at an angle of $30^{\circ}$ with the horizontal with some initial velocity. If its range is $20 \ m$, the maximum height reached by it is: (in meters)

$A$ body of mass $m$ is thrown upwards at an angle $\theta$ with the horizontal with velocity $v$. While rising up,the velocity of the mass after $t$ seconds will be:

$A$ particle of mass $m$ is projected with a velocity $v$ making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is