$A$ ball is thrown at an angle $\theta$ and another ball is thrown at an angle $(90^\circ - \theta)$ with the horizontal from the same point with the same speed $40\,m/s$. The second ball reaches $50\,m$ higher than the first ball. Find their individual heights.

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:

$A$ 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?

The trajectory of a projectile is given by $y = 12x - \frac{3}{4}x^2$. If the horizontal component of velocity is $3 \ m/s$, find the range of the projectile. $(g = 10 \ m/s^2)$ (in $m$)

At what angle of elevation should a projectile be projected with a velocity of $20 \, m/s$ to reach a maximum height of $10 \, m$ (in $^{\circ}$)?

If the potential energy of a projectile at its maximum height is $3/4$ of its initial kinetic energy,then the angle of projection is ...... $^o$.