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  • 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

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:

 

  • Option 1)

    2.5 m

  • Option 2)

    10.3 m

  • Option 3)

    2.8 m

  • Option 4)

    5.1 m

Answers (1)

best_answer

 

Equation of path of a projectile -

y= x \tan \theta\: - \: \frac{gx^{2}}{2u^{2}\cos ^{2}\theta }

it is equation of parabola

g\rightarrow    Acceleratio due to gravity

u\rightarrow    initial velocity

\theta = Angle of projection

 

- wherein

Path followed by a projectile is parabolic is nature.

V_{x}=10\cos\: 60=5m/s

V_{y}=10\cos\: 30=5\sqrt{3}m/s

Velocity after t = 1 sec

V_{x}=5 m/s

1V_{y}1=(10-5\sqrt{3})m/s

Q_{n}=\frac{V^{2}}{R}\Rightarrow R=\frac{Vx^{2}+Vy^{2}}{an}

R=\frac{27+100+75-100\sqrt{3}}{10\cos \Theta }

\Theta =\tan ^{-1}\left ( \frac{10-5\sqrt{3}}{5} \right )=15^{o}

\because R=\frac{100(2-\sqrt{3})}{10\cos 15}=2.8 m

 

 

 


Option 1)

2.5 m

Option 2)

10.3 m

Option 3)

2.8 m

Option 4)

5.1 m

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