The equation of trajectory of a projectile is y-10x- 5 9 x 2- If we assume g-10 ms -2 the range of projectile in metres is-

The correct option is A 18 Given, y=10x 59x^2,g=10m/s^2 So the vertical motion is zero, y=0 0=10x 59x^2⇒ 10x= 59x^2⇒ x=18m ...

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The equation ...

Question

The equation of trajectory of a projectile is $y = 10 x$ , $(\frac{5}{9}) x^{2}$ . If we assume $g = 10 {m s}^{- 2}$ the range of projectile ( in metres) is:

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Similar questions

y = 10 x - ⎧ ⎪ ⎪ ⎪ ⎩ \frac{5}{9} ⎫ ⎪ ⎪ ⎪ ⎭ x^{2}

g = 10 m s^{- 2}

y = 12 x - 3 / 4 x^{2}

g = 10 m s^{- 2}

y = (\frac{x}{\sqrt{3}} - \frac{g x^{2}}{20})

x

y

g = 10 m / s^{2}

y = 10 x - [\frac{5}{9}] x^{2}, (m)

x

y = (\frac{x}{\sqrt{3}} - \frac{g x^{2}}{20})

x

y

g = 10 m / s^{2}

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