Question

A particle is projected with a velocity u = 10 m/s at an angle theta = 60° with the horizontal. the radius of curvature of path of particle at the instant when the velocity vector of the particle becomes perpendicular to initial velocity vector is​

Answer 1

Answer:

80√3

Explanation:

∴v=v

x

i

^

+v

y

j

^

v=i×u

x

cosθ

i

^

+u

x

×cosθ×tan(θ/2)

j

^

v

2

=u

2

cos

2

θ+u

2

cos

2

θ×tan

2

θ/2

v

2

=u

2

cos

2

θ(1+tan

2

θ/2)

v

2

=u

2

cos

2

θ×sec

2

θ/2

∴ Radius of curvature =

gcosθ/2

u

2

cos

2

θ×sec

2

θ/2

=

4×3×10×

3

400×1×4×2

3

3

80