Question

A particle is projected with velocity $\vec{u} = (4 \hat{i} + 3 \hat{j}) m / s$ then the radius of curvature of its trajectory at the highest position is (Take g = 10 m/s²)

Moderate

Solution

At the highest position radius of curvature is given by $g = \frac{{(u \cos α)}^{2}}{ρ}$
$\Rightarrow ρ = \frac{{(u \cos α)}^{2}}{g} = \frac{{(4)}^{2}}{10} m = 1.6 m$

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