$A$ particle is dropped from a height of $20\,m$ above horizontal ground. Due to wind,the particle experiences a constant horizontal acceleration of $6\,m/s^2$. Find the horizontal displacement of the particle when it reaches the ground.

$A$ cannon ball is fired from the top of a $55 \ m$ high cliff with an initial speed of $50 \ m \ s^{-1}$. The speed of the cannon ball while hitting the ground in $m \ s^{-1}$ is (acceleration due to gravity $= 10 \ m \ s^{-2}$):

$A$ body is projected horizontally from the top of a tower of height $180 \ m$ with a velocity of $20 \ ms^{-1}$. If acceleration due to gravity is $10 \ ms^{-2}$,then match the following:

List-$I$ (Kinematic Variable)	List-$II$ (Value)
$A$. Velocity of the body after $1 \ s$ (in $ms^{-1}$)	$I$. $5$
$B$. Horizontal displacement of the body after $1 \ s$ (in $m$)	$II$. $20$
$C$. Vertical displacement of the body after $1 \ s$ (in $m$)	$III$. $10$
$D$. Vertical velocity of the body after $1 \ s$ (in $ms^{-1}$)	$IV$. $22.4$

The correct answer is

$A$ gun can fire shells with maximum speed $v_0$ and the maximum horizontal range that can be achieved is $R = \frac{v_0^2}{g}$. If a target farther away by distance $\Delta x$ (beyond $R$) has to be hit with the same gun,show that it could be achieved by raising the gun to a height at least $h = \Delta x \left[ 1 + \frac{\Delta x}{R} \right]$.

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:

An aeroplane flying $490 \, m$ above ground level at $100 \, m/s$ releases a block. How far on the ground will it strike (in $km$)?