Question

The equation of path of a particle projected obliquely from the ground under gravity is given by y = 2x - x2where x and y are in meter The x and y
axes are an horontal and vertical directions respectively The speed of projection is​

Answer 1

Answer:

The equation of path of a particle projected obliquely from the ground under gravity is given by $y = 2x - x^{2}$ where x and y are in meter The x and y

axes are an horontal and vertical directions respectively The speed of projection is​ 5 m/s.

Explanation:

Let u is the speed of projection and $\alpha$ is the angle of projection.

Then $u_{x}$ = $ucos\alpha$ , $a_{x} = 0$ ; $u_{y} = usin\alpha , a_{y} =-g$

$y = u sin\alpha * t - \frac{1}{2} g t^{2} ................. (i)\\\\x = u cos\alpha * t \\\\or , t = \frac{x}{ucos\alpha }$$..(ii)$

Put (ii) in (i)

$y = usin\alpha * \frac{x}{ucos\alpha } - \frac{1}{2} * g*\frac{x}{ucos\alpha } ^{2}$

$or , y = xtan\alpha - \frac{1}{2} * g * \frac{x^{2} }{u^{2}cos^{2}\alpha }$ $........... (iii)$

$y = 2x - x^{2} .....(iv)$

Comparing $(iii)$ and $(iv)$

$tan\alpha = 2$

$\frac{g}{2u^{2}cos^{2} \alpha } = 1$

if , $tan\alpha = 2$

then $cos \alpha = \frac{1}{\sqrt{1^{2}+2^{2} } } = \frac{1}{\sqrt{5} }$

$\frac{g}{2u^{2}cos^{2}\alpha } = 1\\\\or , \frac{5*10}{2u^{2} } = 1 \\\\\\or , u^{2} = 25.\\\\or , u = 5.$

The equation of path of a particle projected obliquely from the ground under gravity is given by $y = 2x - x^{2}$ where x and y are in meter The x and y

axes are an horontal and vertical directions respectively The speed of projection is​ 5 m/s.

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