The equation of motion of a projectile is $y = 12x - \frac{5}{9}x^2$. The horizontal component of velocity is $3 \ ms^{-1}$. Given that $g = 10 \ ms^{-2}$,the range of the projectile is .......... $m$.

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

$A$ projectile is projected from the ground with an initial velocity $\vec{u} = u_0 \hat{i} + v_0 \hat{j}$. If the acceleration due to gravity $g$ is along the negative $y$-direction,find the maximum displacement in the $x$-direction (horizontal range).

$A$ ball is projected at an angle $45^{\circ}$ with the horizontal. It passes through a wall of height $h$ at a horizontal distance $d_1$ from the point of projection and strikes the ground at a horizontal distance $(d_1 + d_2)$ from the point of projection. Then $h$ is:

It is possible to project a particle with a given velocity in two possible ways so as to make them pass through a point $P$ at a horizontal distance $r$ from the point of projection. If $t_1$ and $t_2$ are times taken to reach this point in two possible ways, then the product $t_1 t_2$ is proportional to

For angles of projection of a projectile at angles $(45^\circ + \theta)$ and $(45^\circ - \theta)$,the horizontal range described by the projectile are in the ratio of: