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Plzz solve this............ ............. ..................

Plzz solve this............
.............  ..................  

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Grade:12th pass

1 Answers

Arun
25750 Points
4 years ago
Dear Aryan
 

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

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