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 horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

$A$ helicopter is flying horizontally with a speed $v$ at an altitude $h$ and has to drop a food packet for a man on the ground. What is the distance of the helicopter from the man when the food packet is dropped?

If the velocity at the maximum height of a projectile projected at an angle of $45^{\circ}$ is $20 \,m \,s^{-1}$, then the maximum height reached by the projectile is (Acceleration due to gravity $=10 \,m \,s^{-2}$ ) (in $\,m$)

If the height of a projectile at a time of $2 \ s$ from the beginning of motion is $60 \ m$,then the time of flight of the projectile is (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $s$)

One second after projection, a projectile is travelling in a direction inclined at $45^{\circ}$ to the horizontal. After two more seconds, it is travelling horizontally. Then the magnitude of the initial velocity of the projectile is $(g = 10 \,ms^{-2})$