The equation of a projectile is $y = 16x - \frac{5x^2}{4}$. The horizontal range is .......... $m$.

The initial velocity of a particle of mass $2\,kg$ is $(4 \hat{i} + 4 \hat{j})\,m/s$. $A$ constant force of $-20 \hat{j}\,N$ is applied on the particle. Initially,the particle was at $(0,0)$. Find the $x$-coordinate of the point where its $y$-coordinate is again zero. $..........\,m$

Match Column-$I$ with Column-$II$.

Column-$I$	Column-$II$
$(1)$ Vertical component of velocity is zero	$(a)$ Tangent to the parabolic path
$(2)$ Linear velocity	$(b)$ Maximum height point of the projectile trajectory

$A$ particle of mass $m$ is thrown with an initial speed $u$ at an angle $\theta$ with the horizontal. Find the torque of its weight about the point of projection when it reaches the highest point.

If the maximum height and range of a projectile are $3 \,m$ and $4 \,m$ respectively,then the velocity of the projectile is (Take $g=10 \,ms^{-2}$)

The velocity of projection of a body is increased by $2 \%$. Other factors remaining unchanged,what will be the percentage change in the maximum height attained? (in $\%$)