an object is projected with a velocity of 20m/s making an angel 45° w

1.	an object is projected with a velocity of 20m/s making an angel 45° with the horizontal.the equation of the trajectory is y=Ax-bx^2 ,where y is height and x is horizontal distance.then A^2/4B is?
Answer» a particle is PROJECTED with a velocity U m/s at an angle θ° with horizontal.then equation of TRAJECTORY is given by here given, θ = 45° and u = 20 m/s and g = 10 m/s² now compare it with y = Ax - Bx² then, A = 1 and B = 1/40 so, A : B = 1 : 1/40 = 40 :1

an object is projected with a velocity of 20m/s making an angel 45° with the horizontal.the equation of the trajectory is y=Ax-bx^2 ,where y is height and x is horizontal distance.then A^2/4B is?

Answer»

a particle is PROJECTED with a velocity U m/s at an angle θ° with horizontal.then equation of TRAJECTORY is given by

here given, θ = 45° and u = 20 m/s and g = 10 m/s²

now compare it with y = Ax - Bx²

then, A = 1 and B = 1/40

so, A : B = 1 : 1/40 = 40 :1

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an object is projected with a velocity of 20m/s making an angel 45° with the horizontal.the equation of the trajectory is y=Ax-bx^2 ,where y is height and x is horizontal distance.then A^2/4B is?​

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an object is projected with a velocity of 20m/s making an angel 45° with the horizontal.the equation of the trajectory is y=Ax-bx^2 ,where y is height and x is horizontal distance.then A^2/4B is?