Question

A boy is standing inside a train moving with a constant velocity of magnitude $10m/s$. He throws a ball vertically up with a speed of $5m/s$ relative to the train. Find the radius of curvature of the path of the ball just at the time of projection. (Take $g=10m/s^2$)

• A. $14m$
• B. $7m$
• C. $5m$
• D. $10m$

Answer 1

The correct option is A $14m$

Initial velocity of the ball in x-direction $=10m/s$
And initial velocity of the ball in y-direction $=5m/s$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)\Rightarrow \tan \theta =\frac{1}{2}$
$\Rightarrow \cos \theta =\frac{2}{\sqrt{3}}$
Since centripetal acceleration, $a_c=\frac{v^2}{r}$,

where r is the radius of curvature)}\)
$\therefore r=\frac{v^2}{g\cos \theta }=\frac{5^2+{10}^2}{10\times \frac{2}{\sqrt{5}}}$
$r=\frac{125\sqrt{5}}{20}m=13.9\approx 14m$