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DLS

  • one year ago

@yrelhan4

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  1. DLS
    • one year ago
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    \[\large x_1=(u_1\cos \theta_1)t_1~,y_1=(u_1 \sin \theta_1)t-\frac{1}{2}g t^2\] \[\large x_2=(u_2\cos \theta_2)t_2~\And~y_2=(u_2 \sin \theta_2)t-\frac{1}{2}g t^2\] The position of one projectile w.r.t another projectile is: \[\large x=x_1-x\] \[\large y=y_1-y_2\] \[\frac{y}{x}=\frac{u_1 \sin \theta_1-u_2 \sin \theta_2}{u_1 \cos \theta_1-u_2 \cos \theta_2}=constant\] \[\LARGE y ~\alpha ~x\] Equation of straight line Thus,Motion of a projectile as observed from another projectile is a straight line

  2. yrelhan4
    • one year ago
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    :O

  3. DLS
    • one year ago
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    :)

  4. DLS
    • one year ago
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    @yrelhan4 bc tujhe kisne medal aur kyun dia -_-

  5. yrelhan4
    • one year ago
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    I'm the boss. \m/ :D

  6. DLS
    • one year ago
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    ._.

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