The horizontal range and the maximum height of a projectile are equal . The angle of projection of the projectile is
${45^o}$
$\theta = {\tan ^{ - 1}}(0.25)$
$\theta = {\tan ^{ - 1}}4$
${60^o}$
The velocity at the maximum height of a projectile is $\frac{\sqrt{3}}{2}$ times its initial velocity of projection $(u)$. Its range on the horizontal plane is .............
A particle is projected in air at some angle to the horizontal, moves along parabola as shown in figure where $x$ and $y$ indicate horizontal and vertical directions respectively. Shown in the diagram, direction of velocity and acceleration at points $A, \,B$ and $C$.
A projectile is fired with a velocity at right angle to the slope which is inclined at an angle $\theta$ with the horizontal. The expression for the range $R$ along the incline is
A stone is thrown at an angle $\theta $ to the horizontal reaches a maximum height $H$. Then the time of flight of stone will be
A particle covers $50\, m$ distance when projected with an initial speed. On the same surface it will cover a distance, when projected with double the initial speed ......... $m$