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  • The equation of trajectory of a projectile is given by  where

The equation of trajectory of a projectile is given by y = -x^2 +10 x , where x and y are in metres and x is along horizontal and y is vertically upward and particles are projected from origin. Then which of the following option is incorrect 

 

Option: 1

Initial velocity of particle is \sqrt{505}m/s


Option: 2

Horizontal range is 10 m 


Option: 3

Maximum height is 25 m 


Option: 4

All of the above


Answers (1)

best_answer

\\ \text{Given :}

 \\ y=- x^2+10x \\ \text{ After arranging the above equation we get }\\ y=10x-x^{2}

Comparing with the Standard equation of trajectory

 y= x\tan \theta - \frac{g x^2}{2u^2\cos ^2 \theta }

\Rightarrow \tan \theta = +10 \Rightarrow \cos \theta = \frac{1}{\sqrt{101}} \\\; \; \; And \ \frac{g }{2 u^2\cos ^2 \theta }= 1 \\\\ \ \; \; u^2 = \; \; \frac{10 }{2\cdot (\frac{1}{101})}= 505

u = \sqrt{505}m /s

R= u^2sin2\theta/{g}= \frac{505 \times 2}{101}= 10 m \\ \text{Maximum Height} ,\left ( H_{max} \right )=\frac{u^2 \sin ^{2}\theta }{2g }

H_{max} = \frac{505 \times \left ( \frac{10}{\sqrt{101}} \right )^2}{2\times 10}= \frac{505\times 100}{20 \times 101}= 25 m

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