At $t = 0$ a projectile is fired from a point $O$(taken as origin) on the ground with a speed of $50\,\, m/s$ at an angle of $53^o$ with the horizontal. It just passes two points $A \& B$ each at height $75 \,\,m$ above horizontal as shown The distance (in metres) of the particle from origin at $t = 2$ sec.
$60\sqrt 2 $
$100$
$60$
$120$
A particle is projected from the ground with an initial speed $\upsilon $ at an angle $\theta $ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is
Two particles are moving along two long straight lines, in the same plane, with the same speed $= 20 \,\,cm/s$. The angle between the two lines is $60^o$, and their intersection point is $O$. At a certain moment, the two particles are located at distances $3\,m$ and $4\,m$ from $O$, and are moving towards $O$. Subsequently, the shortest distance between them will be
The horizontal range and the maximum height of a projectile are equal . The angle of projection of the projectile is
A boy can throw a stone up to a maximum height of $10\ m$. The maximum horizontal distance that the boy can throw the same stone up to will be .......... $m$
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 $g = 10\, ms^{-2}$)