Question

A body is projected at t = 0 with a velocity 10 m second at an angle of 60 degree with the horizontal the radius of curvature of its trajectory at equal 1 second is are neglecting air resistance and taking acceleration due to gravity is equal 10m second time value of rs

Answer 1

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https://www.entrance360.com/engineering/question-stuck-here-help-me-understand-a-body-is-projected-at-t-0-with-a-velocity-10ms-1-at-an-angle-of-600-with-the-horizontal-the-radius-of-curvature-of-its-trajectory-at-t-1s-is-r-neglecting-air-resistance-and-taking-acceleration-due-to-gravity-g-10-ms-2-the-value-of-r-is/

Answer 2

The value of R is 2.8 m.

The body is following the path of a projectile.

velocity V = 10m/s

So velocity across x-direction,

Vx = V cos Φ, where Φ = 60°

So, Vx = 10 cos 60° = 5 m/s                                [cos 60° = 0.5]

Again, velocity across y-direction,

Vy = V sinΦ = V sin 60° = 5√3 m/s                      [sin 60° = √3/2]

After 1s, Vy changes to Vy - V

So, Vy = (5√3 - 10) m/s

Magnitude of Vy = | (5√3 - 10) | m/s = (10 - 5√3) m/s

Again, an = V²/R, and V² = (Vx² + Vy²)

So, R = (Vx² + Vy²)/an,

where an = V cos ∅

R = (25 + 100 + 75 - 100√3)/10 cos∅       ...(1)

or, tan ∅ = (10 - 5√3)/5 = (2 - √3)

or, ∅ = tan inverse (2 - √3) = 15°

Putting the value of ∅ in eqn.(1), we get:

R = (25 + 100 + 75 - 100√3)/10 cos 15°

  = 2.8 m

which is the required value.