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(D) Given: Initial velocity $u = 20 \ m/s$,angle of projection $	heta = 45^{\circ}$,and $g = 10 \ m/s^2$.The standard equation of the trajectory of a

Vedclass · Accepted Answer

$40:1$

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(D) Given: Initial velocity $u = 20 \ m/s$,angle of projection $	heta = 45^{\circ}$,and $g = 10 \ m/s^2$.The standard equation of the trajectory of a projectile is given by $h = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Comparing this with the given equation $h = Ax - Bx^2$,we get:$A = 	an 	heta = 	an 45^{\circ} = 1$.$B = \frac{g}{2u^2 \cos^2 	heta} = \frac{10}{2 	imes (20)^2 	imes (\cos 45^{\circ})^2} = \frac{10}{2 	imes 400 	imes (1/\sqrt{2})^2} = \frac{10}{800 	imes 1/2} = \frac{10}{400} = \frac{1}{40}$.Now,the ratio $A:B$ is calculated as:$\frac{A}{B} = \frac{1}{1/40} = 40$.Therefore,the ratio $A:B$ is $40:1$.

An object is projected with a velocity of $20 \ m/s$ making an angle of $45^{\circ}$ with the horizontal. The equation for the trajectory is $h = Ax - Bx^2$,where $h$ is the height,$x$ is the horizontal distance,and $A$ and $B$ are constants. The ratio $A:B$ is $(g = 10 \ m/s^2)$.

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