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:

Q. The maximum height attained by a projectile when thrown at an angle $0$ with the horizontal is found to be half the horizontal is found to be half the horizontal range Then is equal to:

Q.

A stone projected with a velocity u at an angle $\theta$ with the horizontal reaches maximum height $H_1$. When it is projected with velocity u at an angle $(\frac{\pi }{2}-\theta )$ with the horizontal, it reaches maximum height $H_2$.
The relation between the horizontal range R of the projectile, $H_1$ and $H_2$ is

Q. The A stone projected with a velocity $u$ at an angle $\Theta$ with the horizontal reaches maximum height $H_1$. When it is projected with velocity $u$ at an angle $(\frac{\pi }{2}-\Theta )$ with the horizontal, it reaches maximum height $H_2$. The relation between the horizontal range $R$ of the projectile $H_1$ and $H_2$ is,

Q. A projectile is projected from ground with velocity $v_0$ at an angle $\theta$ with horizontal. What is the velocity of projectile when its vertical displacement is equal to half of the maximum height attained?

Q. If angles of projection are $(\frac{\pi }{4}+\theta )$ and $(\frac{\pi }{4}-\theta )$ where $\theta <\frac{\pi }{4}$, then the ratio of horizontal range described by the projectile is (speed is same):