Question

A stone is thrown horizontally with a velocity $10m/s$. The radius of curvature of its trajectory in $3$ second after the motion began is $100\sqrt{2a}m$. Disregard the resistance of air. Determine $a$.

Answer 1

Answer: 5.00

$v_y=0+gt=10\times 3=30m/\sec$
$v_x=10m/\sec$
$\therefore$ net velocity $=v_{\text{net }}=\sqrt{v_x^2+v_y^2}$
$=\sqrt{(10{)}^2+(30{)}^2}=10\sqrt{10}m/\sec$
The acceleration, normal to tangential velocity
$=g\cos \theta =g\times \frac{v_x}{v_{\text{net }}}=\frac{10\times 10}{10\sqrt{10}}=\sqrt{10}$
So, radius of curvature
$=\frac{v_{\text{net }}^2}{g\cos \theta }=\frac{1000}{\sqrt{10}}$
$=100\sqrt{10}m$