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Question

# What is the speed of a projectile at a point on its path where it makes an angle of 45∘ with the horizontal? It is given that the radius of curvature (R) of particle at this point is 50√2 m. Take g=10 m/s2.

A
2505 m/s
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B
102 m/s
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C
105 m/s
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D
2502 m/s
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Solution

## The correct option is C 10√5 m/sFor finding the radius of curvature we apply the concept of centripetal acceleration, where we resolve the component of acceleration (g) of particle towards the centre of circle with radius of curvature (R) as radius. ac=v2r..........(1) From Eq.(1) replacing ac→gcosθ, r→R: R=v2gcosθ ⇒v2=R×gcosθ ⇒v2=50√2×10cos45∘ ∴ v=10√5 m/s

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