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(D) The standard equation of projectile motion is given by $h = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$.Comparing this with the given equat

Vedclass · Accepted Answer

$40:1$

Vedclass · Answer

(D) The standard equation of projectile motion is given by $h = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$.Comparing this with the given equation $h = Ax - Bx^2$,we identify:$A = 	an 	heta$$B = \frac{g}{2 u^2 \cos^2 	heta}$Given $	heta = 45^\circ$,$u = 20 \ m/s$,and $g = 10 \ m/s^2$:$A = 	an 45^\circ = 1$$B = \frac{10}{2 	imes (20)^2 	imes \cos^2 45^\circ} = \frac{10}{2 	imes 400 	imes (1/\sqrt{2})^2} = \frac{10}{800 	imes 0.5} = \frac{10}{400} = \frac{1}{40}$Therefore,the ratio $A : B = 1 : (1/40) = 40 : 1$.

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 height,$x$ is horizontal distance,and $A$ and $B$ are constants. The ratio $A : B$ is $(g = 10 \ m/s^2)$.

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