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  • Equation of trajectory of a projectile is given by

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 is projected from origin .Then which of the following option is incorrect 

 

A.

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

B.

horizontal range is 10 m 

C.

Maximum range is 10 m 

D.

Angle of projection with horizontal is \tan ^{-1} (-1 )

505m/s

Answers (1)

@Harshit

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.

 

 y=- x^2+10x

Compare this with 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

Maximum height H = \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

 

 

 

Posted by

Safeer PP

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