The horizontal and vertical displacements $x$ and $y$ of a projectile at a given time $t$ are given by $x = 6t \text{ m}$ and $y = 8t - 5t^2 \text{ m}$. The range of the projectile in metres is:

$A$ particle is projected with a velocity of $30\,m/s$ at an angle of $\theta_0 = \tan^{-1}(3/4)$. After $1\,s$,the particle is moving at an angle $\theta$ to the horizontal. What is the value of $\tan\theta$? (Take $g = 10\,m/s^2$)

Two bodies are projected with the same velocity. If one is projected at an angle of $30^\circ$ and the other at an angle of $60^\circ$ to the horizontal,the ratio of the maximum heights reached is

Assertion $(A)$: The range of a projectile is maximum when the angle of projection is $45^{\circ}$.
Reason $(R)$: The range of a projectile depends only on the angle of projection.

$A$ bomb is dropped by an airplane flying horizontally with a velocity of $200 \text{ km/hr}$ at a height of $980 \text{ m}$. At the time of dropping the bomb,the horizontal distance of the airplane from the target on the ground to hit it directly is (given $g = 9.8 \text{ m/s}^2$):

$A$ stone is projected from the ground with velocity $50 \, m/s$ at an angle of $30^o$. It crosses a wall after $3 \, s$. How far beyond the wall will the stone strike the ground? $(g = 10 \, m/s^2)$