$A$ projectile is thrown with a velocity of $50 \, ms^{-1}$ at an angle of $53^o$ with the horizontal. The equation of the trajectory is given by:

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.

$A$ missile is fired for maximum range at your town from a place $100\, km$ away from you. If the missile is fired for maximum range,what is the maximum height reached by the missile? (Take $g = 10\, m/s^2$)

$A$ projectile is thrown at an angle $\theta$ with the horizontal and its range is $R_1$. It is then thrown at an angle $\theta$ with the vertical and the range is $R_2$. Then:

$A$ ball is projected from a building of height $10 \ m$ with a velocity of $10 \ m/s$ at an angle of $30^\circ$ with the horizontal. What is the horizontal distance covered by the ball when it reaches the same height of $10 \ m$ again (in $m$)? $(g = 10 \ m/s^2, \sin 30^\circ = 1/2, \cos 30^\circ = \sqrt{3}/2)$

$A$ particle is projected at an angle of $30^{\circ}$ from the horizontal at a speed of $60 \; m/s$. The height traversed by the particle in the first second is $h_0$ and the height traversed in the last second before it reaches the maximum height is $h_1$. The ratio $h_0 : h_1$ is . . . . . . . [Take $g = 10 \; m/s^2$]