$A$ 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 $g = 10\,m/s^2$)

An object is thrown along a direction inclined at an angle of $45^\circ$ with the horizontal direction. The horizontal range of the particle is equal to

$A$ particle is projected with velocity $2 \sqrt{gh}$ at an angle $60^{\circ}$ to the horizontal so that it just clears two walls of equal height $h$ which are at a distance $2h$ from each other. The time taken by the particle to travel between these two walls is

The trajectory of a projectile in a vertical plane is $y = \alpha x - \beta x^2$,where $\alpha$ and $\beta$ are constants and $x$ and $y$ are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection $\theta$ and the maximum height attained $H$ are respectively given by:

$A$ missile is fired for maximum range with an initial velocity of $20\; m/s$. If $g = 10\; m/s^2$,the range of the missile is ...... $m$.

Define a projectile and derive the equation of its trajectory: $y = (\tan \theta_0)x - \frac{g}{2(v_0 \cos \theta_0)^2}x^2$.