The ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is $x$. The angle of projection with the horizontal is

$A$ 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 with a velocity of $10 \ ms^{-1}$ at an angle of $60^{\circ}$ with the vertical direction. Its speed at the highest point of its trajectory will be $............... \ ms^{-1}$.

$A$ projectile is given an initial velocity of $(\hat{i}+2 \hat{j}) \text{ ms}^{-1}$. The equation of its path is $(g=10 \text{ ms}^{-2})$

If $R$ and $H$ are the horizontal range and maximum height attained by a projectile,then its speed of projection is ..........

$A$ 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: