A projectile has the same range $R$ for two angles of projection. If $T_1$ and $T_2$ be the times of flight in the two cases, then $R$ is
$T_1T_2g$
$\frac{T_1T_2g}{2}$
$(T_1^2 + T_2^2)g$
$\frac{(T_1^2 + T_2^2)}{2}g$
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 $ ?
An object is projected from ground with speed $u$ at angle $\theta$ with horizontal. the radius of curvature of its trajectory at maximum height from ground is ..........
An object is projected with a velocity of $20 m/s$ making an angle of $45^o$ with horizontal. The equation for the trajectory is $h = Ax -Bx^2$ where $h$ is height, $x$ is horizontal distance, $A$ and $B$ are constants. The ratio $A : B$ is ($g = 10 ms^{-2}$)
If time of flight of a projectile is $10$ seconds. Range is $500$ meters. The maximum height attained by it will be ......... $m$
Galileo writes that for angles of projection of a projectile at angles $(45 + \theta )$ and $(45 - \theta )$, the horizontal ranges described by the projectile are in the ratio of (if $\theta \le 45)$