The range of a projectile when launched at angle $\theta$ is same as when launched at angle $2\theta$. What is the value of $\theta$ (in $^{\circ}$)?

The path of a projectile is given by the equation $y = ax - bx^2$, where $a$ and $b$ are constants, and $x$ and $y$ are the horizontal and vertical distances of the projectile from the point of projection, respectively. The maximum height attained by the projectile and the angle of projection are respectively:

The velocity of a projectile is $(6\hat{i} + 8\hat{j}) \ m/s$. Its horizontal range is ........ $m$.

$A$ ball thrown by one player reaches another player in $2 \text{ s}$. The maximum height attained by the ball above the point of projection will be $.... \text{ m}$.

$A$ projectile is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal direction with a velocity of $147 \ m/s$. Then the time after which its inclination with the horizontal is $45^{\circ}$ is ......... $s$.

$A$ projectile is given an initial velocity of $(\hat i + 2\hat j) \ ms^{-1}$,where $\hat i$ is along the ground and $\hat j$ is along the vertical. If $g = 10 \ m/s^2$,the equation of its trajectory is