The horizontal range and maximum height attained by a projectile are $R$ and $H$, respectively. If a constant horizontal acceleration $a=g / 4$ is imparted to the projectile due to wind, then its horizontal range and maximum height will be

A particle moves in a plane with constant acceleration in a direction different from the initial velocity. The path of the particle will be

The position of a projectile launched from the origin at $t = 0$ is given by $\vec r = \left( {40\hat i + 50\hat j} \right)\,m$ at $t = 2\,s$. If the projectile was launched at an angle $\theta$ from the horizontal, then $\theta$ is (take $g = 10\, ms^{-2}$)

An object is projected at an angle of $45^°$ with the horizontal. The horizontal range and the maximum height reached will be in the ratio.

A particle is projected from the ground with velocity $u$ at angle $\theta$ with horizontal. The horizontal range, maximum height and time of flight are $R, H$ and $T$ respectively. They are given by $R = \frac{{{u^2}\sin 2\theta }}{g}$, $H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ and $T = \frac{{2u\sin \theta }}{g}$  Now keeping $u $ as fixed, $\theta$ is varied from $30^o$ to $60^o$. Then,