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Grade 12th passMechanics

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

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Aryan

7 Years agoGrade 12th pass

1 Answer

Arun

7 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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