$A$ ball is thrown upward from the top of a building at an angle of $30^{\circ}$ to the horizontal with an initial speed of $15 \,ms^{-1}$. If the ball hits the ground after $3 \,s$, then the height of the building is (acceleration due to gravity $= 10 \,ms^{-2}$): (in $\,m$)

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 $m$)

$A$ particle is dropped from a height of $20\,m$ onto horizontal ground. There is wind blowing,due to which the horizontal acceleration of the particle becomes $6\,ms^{-2}$. Find the horizontal displacement of the particle until it reaches the ground (in $m$).

$A$ 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:

From the top of a tower of height $40 \ m,$ a ball is projected upwards with a speed of $20 \ m/s$ at an angle $30^{\circ}$ to the horizontal. The ball will hit the ground in time ........ $s$ (Take $g = 10 \ m/s^2$)

Two particles in the same vertical plane are thrown to strike at the same time. One is thrown from the ground and the other from a height $h$ vertically above it. The ground particle is thrown obliquely with speed $u$ at an angle $\theta$ and achieves a maximum height $H$. The second particle is thrown horizontally with the same speed $u$. What is the maximum height $h$ such that the two particles strike each other in the air?