$A$ 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 thrown with speed $40 \, m/s$ at an angle $\theta$ from the horizontal. It is found that the projectile is at the same height at $1 \, s$ and $3 \, s$. What is the angle of projection?

$A$ particle is projected with an angle of projection $\theta$ to the horizontal. The trajectory passes through the points $(P, Q)$ and $(Q, P)$ referred to horizontal and vertical axes ($x$-axis and $y$-axis respectively). The angle of projection $\theta$ is given by:

For an object projected from the ground with speed $u$,the horizontal range is two times the maximum height attained by it. The horizontal range of the object is ..........

Galileo,in his book $Two \text{ } new \text{ } sciences$,stated that "for elevations which exceed or fall short of $45^{\circ}$ by equal amounts,the ranges are equal". Prove this statement.

The equation of the trajectory of a ball projected at an angle $\theta$ with the horizontal is given as $y = x - \frac{gx^2}{2}$. The initial velocity of the ball is $[\text{Given} : \tan 45^{\circ} = 1, \cos 45^{\circ} = \frac{1}{\sqrt{2}}]$