Question

A stone is projected with speed Vo at angle of projection 53° with the horizontal. At the instant the velocity of the stone makes angle 37° with the horizontal, the radius of trajectory of its path is​

Answer 1

In the fig., since the horizontal velocity is constant throughout the motion,

$\sf{\longrightarrow v_0\cos\theta=v\cos\alpha}$

$\sf{\longrightarrow v=\dfrac{v_0\cos\theta}{\cos\alpha}\quad\quad\dots(1)}$

From FBD of the projectile at the point where it makes an angle $\alpha$ with horizontal, we get its weight component is balanced by centripetal force, i.e.,

$\sf{\longrightarrow \dfrac{mv^2}{R}=mg\cos\alpha}$

$\sf{\longrightarrow R=\dfrac{v^2}{g\cos\alpha}}$

From (1),

$\sf{\longrightarrow R=\dfrac{(v_0)^2\cos^2\theta}{g\cos^3\alpha}}$

where $\sf{R}$ is the radius of curvature of the projectile at that point.

In this question,

• $\sf{\theta=53^o}$

• $\sf{\alpha=37^o}$

• $\sf{g=10\ m\,s^{-2}}$

Then,

$\sf{\longrightarrow R=\dfrac{(v_0)^2\cos^253^o}{g\cos^337^o}}$

$\sf{\longrightarrow R=\dfrac{(v_0)^2\cdot\dfrac{9}{25}}{10\cdot\dfrac{64}{125}}}$

$\sf{\longrightarrow\underline{\underline{R=\dfrac{9}{128}\,(v_0)^2}}}$

Answer 2

9/128 is your answer ..