Equation of trajectory of a projectile is given as y=x−x240. The radius of curvature at highest point of its trajectory is [Take, g=10 m/s2 ]

y=x−x240y=xtanθ−gx22u2cos2θtanθ=1 ; θ=450102u2.12=140u=20 m/sRadius of curvature at highest point =u2cos2θg=4000×1210=20 m

Equation of trajectory of a projectile is given as y=x−x240. The radius of curvature at highest point of its trajectory is [Take, g=10 m/s2 ]

y=x−x240y=xtanθ−gx22u2cos2θtanθ=1 ; θ=450102u2.12=140u=20 m/sRadius of curvature at highest point =u2cos2θg=4000×1210=20 m

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Equation of trajectory of a projectile is given as $y = x - \frac{x^{2}}{40} .$ The radius of curvature at highest point of its trajectory is [Take, $g = 10 m / s^{2}$ ]

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5 m

40 m

10 m

20 m

answer is B.

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Detailed Solution

$y = x - \frac{x^{2}}{40}$

$y = x \tan θ - \frac{g x^{2}}{2 u^{2} \cos^{2} θ}$

$\tan θ = 1; θ = 45^{0}$

$\frac{10}{2 u^{2} . \frac{1}{2}} = \frac{1}{40}$

$u = 20 m / s$

Radius of curvature at highest point $= \frac{u^{2} \cos^{2} θ}{g} = \frac{4000 \times \frac{1}{2}}{10} = 20 m$

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Equation of trajectory of a projectile is given as y=x−x240. The radius of curvature at highest point of its trajectory is [Take, g=10 m/s2 ]

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Equation of trajectory of a projectile is given as $y = x - \frac{x^{2}}{40} .$ The radius of curvature at highest point of its trajectory is [Take, $g = 10 m / s^{2}$ ]