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 ...... $^o$ $(g = 10\, m/s^2)$.

Two balls are thrown horizontally from the top of a tower with velocities $v_1$ and $v_2$ in opposite directions at the same time. After how much time will the angle between the velocities of the balls become $90^o$?

$A$ ball of mass $m=1 \, kg$ is thrown from the top of a building with initial velocity $v=(20 \, m/s) \hat{i} + (24 \, m/s) \hat{j}$ at time $t=0$. The change in the potential energy of the ball between $t=0$ and $t=6 \, s$, if the ball does not hit the ground, is (assume $g=10 \, m/s^2$): (in $ \, J$)

$A$ 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

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 $g = 10 \ m/s^2$)

$A$ body is projected horizontally with a velocity of $4\,m/s$ from the top of a high tower. The velocity of the body after $0.7\,s$ is nearly $.....\,m/s$ (take $g=10\,m/s^2$).