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Projectile Motion Physics Motion in Plane Physics Class 11th

The trajectory of a projectile in an vertical plane is y = ax - bx², where a and b are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection q and the maximum height attained H are respectively given by:

Option 1 -

$t a n^{- 1} α, \frac{4 α^{2}}{β}$

Option 2 -

$t a n^{- 1} α, \frac{α^{2}}{4 β}$

Option 3 -

$t a n^{- 1} β, \frac{α^{2}}{2 β}$

Option 4 -

$t a n^{- 1} (\frac{β}{α}), \frac{α^{2}}{β}$

2 Views | Posted 2 weeks ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 weeks ago

Correct Option - 4

Detailed Solution:

$y = α x - β x^{2}$

$\Rightarrow \frac{d y}{d x} = α - 2 β x = 0$

$\Rightarrow x = \frac{α}{2 β}$

$y_{m a x} = α \times \frac{α}{2 β} - β \times {(\frac{α}{2 β})}^{2} = \frac{α^{2}}{4 β}$

Range = $2 x = \frac{α}{β} = \frac{2 u^{2} s i n θ . c o s θ}{g}$

On comparing with

y = x tan θ - $\frac{g x^{2}}{2 u^{2} c o s^{2} θ}$

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The trajectory of a projectile in an vertical plane is y = ax - bx², where a and b are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection q and the maximum height attained H are respectively given by:

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The trajectory of a projectile in an vertical plane is y = ax - bx2, where a and b are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection q and the maximum height attained H are respectively given by:

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The trajectory of a projectile in an vertical plane is y = ax - bx², where a and b are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection q and the maximum height attained H are respectively given by: