$A$ projectile covers double the range compared to its maximum height attained. The angle of projection is

The equation for the trajectory of a projectile is $y = \left( \frac{x}{\sqrt{3}} - \frac{x^2}{60} \right) \text{ m}$. The velocity of projection of the projectile is (Acceleration due to gravity $g = 10 \text{ m s}^{-2}$) (in $\text{ m s}^{-1}$)

$A$ ball having kinetic energy $KE$ is projected at an angle of $60^{\circ}$ from the horizontal. What will be the kinetic energy of the ball at the highest point of its flight?

$A$ ball is projected from the ground into the air. At a height of $5 \text{ m}$, its velocity is $\vec{v} = (5 \hat{i} + 5 \hat{j}) \text{ m s}^{-1}$. The maximum height reached by the ball is (Acceleration due to gravity $g = 10 \text{ m s}^{-2}$): (in $\text{ m}$)

$A$ ball is dropped from a high tower,and another ball is thrown horizontally from the same position. Which ball will reach the ground first?

$A$ ball is projected from the ground with a velocity $V$ at an angle $\theta$ to the vertical. On its path,it makes an elastic collision with a vertical wall and returns to the ground. The total time of flight of the ball is: