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  • Projectile motion when projected horizontally with a speed of 20 m/s, Its angle made with the direction perpendicular to the ground  after  <img alt="t=2\sqr

Projectile motion when projected horizontally with a speed of 20 m/s, Its angle made with the direction perpendicular to the ground  after  t=2\sqrt{5} \ s is ?

Option: 1

{\mathrm{tan}}^{-1}\left(\frac{1}{\sqrt3}\right)


Option: 2

\ \tan^{-1}\sqrt5


Option: 3

\tan^{-1}\sqrt3


Option: 4

\tan^{-1}\left(\frac{1}{\sqrt5}\right)


Answers (1)

best_answer

Answer: (4)  \tan^{-1}\left(\frac{1}{\sqrt5}\right)

Explanation: 

tan\theta=\frac{v_x}{v_y}

Let,

     v_x=u   and      v_y=gt

    tan\theta\ =\frac{u}{gt}                                             

Or, \frac{20}{10\times2\sqrt5}                                           

Or,   tan\theta=\frac{1}{\sqrt5}                                

Or,   \theta\ =\ \tan^{-1}\left(\frac{1}{\sqrt5}\right)               

Posted by

avinash.dongre

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