At a height $0.4\, m$ from the ground,the velocity of a projectile in vector form is $\vec v = (6\hat i + 2\hat j)\,m/s$. The angle of projection is ...... $^o$ $(g = 10\, m/s^2)$.

$A$ 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

The figure shows a projectile thrown with a speed $u = 20 \, m/s$ at an angle of $30^{\circ}$ with the horizontal from the top of a building $40 \, m$ high. The horizontal range of the projectile is ........... $m$.

$A$ cricket bowler releases the ball in two different ways:
$(a)$ giving it only horizontal velocity,and
$(b)$ giving it horizontal velocity and a small downward velocity.
The speed $V_s$ at the time of release is the same. Both are released at a height $H$ from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.

$A$ 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.

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 of elevation of $30^{\circ}$. The ratio of the total time taken by the ball to hit the ground to its time of flight (time taken to come back to the same elevation) is (take $g=10\,m/s^2$).