$A$ body projected with some velocity at an angle $45^{\circ}$ with the horizontal from the origin in $XY$-plane passes through a point at $(4, 3) \ m$. Its horizontal range is (in $m$)

$A$ particle is projected from a horizontal plane ($x-z$ plane) such that its velocity vector at time $t$ is given by $\vec{V} = a\hat{i} + (b - ct)\hat{j}$. Its range on the horizontal plane is given by:

Two projectiles,one fired from the surface of the Earth with velocity $10 \, m/s$ and the other fired from the surface of another planet with initial speed $5 \, m/s$,trace identical trajectories. The value of the acceleration due to gravity on the planet is ......... $m/s^2$.

Assertion $(A)$: The velocity of a projectile at a point on its trajectory is equal to the slope at that point.
Reason $(R)$: The velocity vector at a point is always along the tangent to the trajectory at that point.

If the initial velocity in the horizontal direction of a projectile is the unit vector $\hat{i}$ and the equation of the trajectory is $y = 5x(1 - x)$,find the $y$-component vector of the initial velocity. (Take $g = 10\,m/s^2$) (in $,\hat{j}$)

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