The maximum height attained by a projectile when thrown at an angle $\theta$ with the horizontal is found to be half the horizontal range. Then $\theta$ is equal to

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:

$A$ piece of marble is projected from the Earth's surface with a velocity of $19.6 \sqrt{2} \ m/s$ at an angle of $45^{\circ}$. $2 \ s$ later,its velocity makes an angle $\alpha$ with the horizontal,where $\alpha$ is $..........$ (in $^{\circ}$)

$A$ ball is thrown at $30^{\circ}$ with the horizontal from the top of a roof $20 \,m$ high with a speed of $13 \,ms^{-1}$. At what distance from the throwing point will the ball, once again, be at a height of $20 \,m$ from the ground (in $\,m$)? $(g = 10 \,ms^{-2})$

$A$ body is projected at an angle of $30^{\circ}$ with the horizontal with momentum $p$. At its highest point,the magnitude of the momentum is

$A$ stone is projected from the ground at $t = 0$. At the time of projection,the horizontal and vertical components of velocity are $10\, m/s$ and $20\, m/s$ respectively. Find the time at which the magnitudes of tangential and normal acceleration will be equal $(g = 10\, m/s^2)$ [neglect air friction].