An aeroplane is flying in a horizontal direction with a velocity of $540 \ km/h$ at a height of $1960 \ m$. When it is vertically above the point $A$ on the ground,a body is dropped from it. The body strikes the ground at point $B$. The distance $AB$ is equal to $(g = 9.8 \ m/s^2)$ (in $m$)

Shots are fired from the top of a tower and from its bottom simultaneously at angles $30^o$ and $60^o$ as shown. If the horizontal distance of the point of collision is at a distance $a$ from the tower,then the height of the tower $h$ is:

$A$ boy playing on the roof of a $10 \, m$ high building throws a ball with a speed of $10 \, m/s$ at an angle $30^{\circ}$ with the horizontal. How far from the throwing point will the ball be at the height of $10 \, m$ from the ground (in $, m$)? $(g = 10 \, m/s^2, \sin 30^{\circ} = 1/2, \cos 30^{\circ} = \sqrt{3}/2)$

$A$ body is projected from the top of a tower with a velocity $\overrightarrow{u} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \text{ m/s}$, where $\hat{i}, \hat{j},$ and $\hat{k}$ are unit vectors along east, north, and vertically upwards, respectively. If the height of the tower is $30 \text{ m}$, find the horizontal range of the body on the ground. (Take $g = 10 \text{ m/s}^2$) (in $\text{ m}$)

Projectiles $A$ and $B$ are thrown at angles of $45^{\circ}$ and $60^{\circ}$ with the vertical,respectively,from the top of a $400 \ m$ high tower. If their ranges and times of flight are the same,the ratio of their speeds of projection $v_A : v_B$ is:

$A$ ball of mass $0.2 \ kg$ is thrown from a height of $1 \ m$ with an initial velocity of $\sqrt{10} \ m/s$ at an angle of $45^{\circ}$ with the horizontal. Assuming acceleration due to gravity $g = 10 \ m/s^2$, the modulus of the momentum increment during the total time of motion in $kg \cdot m/s$ is: