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Grade 11Most Scoring Topics in IIT JEE

a particle is projected horizontally from the top of a tower with velocity 5m/s.the radius of the curvature of the path of the particle after 0.5 sec is

Profile image of ananyanair1107
8 Years agoGrade 11
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1 Answer

Profile image of garima soni
8 Years ago
v2/perpendicular acceleration is the formula for radius of carvature. 
v=5m/s in horizontal always.
v in vertical is .5.10=5m/s
so speed=5root2 m/s
so r=(5root2)2/10=5m

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