The maximum height attained by a projectile is increased by $10\,\%$ by increasing its speed of projection,without changing the angle of projection. The percentage increase in the horizontal range will be $...........\,\%$

$A$ particle is projected with an angle of projection $\theta$ to the horizontal. The trajectory passes through the points $(P, Q)$ and $(Q, P)$ referred to horizontal and vertical axes ($x$-axis and $y$-axis respectively). The angle of projection $\theta$ is given by:

$A$ projectile is thrown with an initial velocity of $(a \hat{i} + b \hat{j}) \text{ m s}^{-1}$. If the range of the projectile is twice the maximum height reached by it,then

The maximum vertical height to which a man can throw a ball is $136\,m$. The maximum horizontal distance up to which he can throw the same ball is $.....\,m$.

In the given figure for a projectile,find the relation for $y$ in terms of $x_1, x_2$ and $\theta$.

If the velocity of a projectile at its maximum height is $\frac{1}{\sqrt{2}}$ times its initial velocity,what is its range?