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The Equation for the Path of Projectile

40 the equati...

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40 the equation of trajectory of a projectile projected with a velocity of 10m/satanangle45degreewithverticalisgivenby(g=10m/s)(a)y^2=10(x+y)(b)y^2=10(x-y)(c)x^2=10(x+y)(d)x^2=10(x-y)

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Q. The equation of trajectory of a projectile is

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

g = 10 m s^{- 2}

, the range of projectile (in metres) is:

Q. The equation of trajectory of a projectile is

y = 10 x

(\frac{5}{9}) x^{2}

g = 10 {m s}^{- 2}

the range of projectile ( in metres) is:

Q. Equation of trajectory of a projectile is given by

y = - x^{2} + 10 x

where x and y are in meters and x is along horizontal and y is vertically upward and particle is projected from origin. Then which of the following options is/are correct.

(g = 10 m / s^{2})

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Q. The equation of trajectory of a projectile is

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

. The maximum height is reached when

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Q. The equation of a projectile is

y = (10 x - \frac{5 x^{2}}{4}) m

. The maximum height achieved by the projectile is

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The Equation for the Path of Projectile

Standard XII Physics

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