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

Laws of Motion

Solution:

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$v _{ y }=0+ gt =10 \times 3=30 \,m / \sec$
$v _{ x }=10\, m / \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
$=\frac{v_{\text {net }}^{2}}{g \cos \theta}=\frac{1000}{\sqrt{10}}$
$=100 \sqrt{10}\, m$