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# Need explanation for: - Kinematics - JEE Main-3

The trajectory of a projectile near the surface of the earth is given as $y=2x-9x^{2}$. If it were launched at an angle $\theta_{0}$ with speed $v_{0}$ then $\left ( g=10ms^{-2} \right )$:

• Option 1)

$\theta_{0}=\sin^{-1}\left ( \frac{2}{\sqrt{5}} \right )$ and $v_{0}=\frac{3}{5}ms^{-1}$

• Option 2)

$\theta_{0}=\cos^{-1}\left ( \frac{2}{\sqrt{5}} \right )$ and $v_{0}=\frac{3}{5}ms^{-1}$

• Option 3)

$\theta_{0}=\cos^{-1}\left ( \frac{1}{\sqrt{5}} \right )$ and $v_{0}=\frac{5}{3}ms^{-1}$

• Option 4)

$\theta_{0}=\sin^{-1}\left ( \frac{1}{\sqrt{5}} \right )$ and $v_{0}=\frac{5}{3}ms^{-1}$

Answers (1)
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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.

Given,      $y=2x-9x^{2}$

General Eqn.  $\rightarrow y= x\tan\theta-\frac{1}{2}\frac{gx^{2}}{v^{2}\cos^{2}\theta}$

$\Rightarrow \tan\theta=2$

$\Rightarrow \theta =\sin^{-1}\left ( \frac{2}{\sqrt{5}} \right )=\cos^{-1}\left ( \frac{1}{\sqrt{5}} \right )$

$\frac{g}{2v^{2}\cos^{2}\theta}=9$

$\Rightarrow \frac{g}{2v^{2}\times\frac{1}{5}}=9$

$\Rightarrow \frac{10\times5}{2\times9}=v^{2}$

$\Rightarrow v^{2}=\frac{25}{9}$

$v=\frac{5}{3}m/s$

Option 1)

$\theta_{0}=\sin^{-1}\left ( \frac{2}{\sqrt{5}} \right )$ and $v_{0}=\frac{3}{5}ms^{-1}$

Option 2)

$\theta_{0}=\cos^{-1}\left ( \frac{2}{\sqrt{5}} \right )$ and $v_{0}=\frac{3}{5}ms^{-1}$

Option 3)

$\theta_{0}=\cos^{-1}\left ( \frac{1}{\sqrt{5}} \right )$ and $v_{0}=\frac{5}{3}ms^{-1}$

Option 4)

$\theta_{0}=\sin^{-1}\left ( \frac{1}{\sqrt{5}} \right )$ and $v_{0}=\frac{5}{3}ms^{-1}$

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