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Q.

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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a

5 m

b

40 m

c

10 m

d

20 m

answer is B.

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Detailed Solution

y=x−x240

y=xtanθ−gx22u2cos2θ

tanθ=1  ;   θ=450

102u2.12=140

u=20 m/s

Radius of curvature at highest point =u2cos2θg=4000×1210=20 m

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