If $R$ and $H$ are the horizontal range and maximum height attained by a projectile, than its speed of projection is ..........
$\sqrt{2 g R+\frac{4 R^2}{g H}}$
$\sqrt{2 g H+\frac{R^2 g}{8 H}}$
$\sqrt{2 g H+\frac{8 H}{R g}}$
$\sqrt{2 g H+\frac{R^2}{H}}$
The maximum vertical height to which a man can throw a ball is $136\,m$. The maximum horizontal distance upto which he can throw the same ball is $.....\,m$
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.
A player throws a ball that reaches to the another player in $4\,s$. If the height of each player is $1.5\,m$, the maximum height attained by the ball from the ground level is .......... $m$
If a stone is to hit at a point which is at a distance $d$ away and at a height $h$ above the point from where the stone starts, then what is the value of initial speed $u$ if the stone is launched at an angle $\theta $ ?
For a given velocity, a projectile has the same range $R$ for two angles of projection if $t_1$ and $t_2$ are the times of flight in the two cases then