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(B) The maximum height $H$ of a projectile is given by the formula $H = \frac{u^2 \sin^2 \alpha}{2g}$,where $u$ is the initial speed and $\alpha$ is t

Vedclass · Accepted Answer

$	an^2 	heta : 1$

Vedclass · Answer

(B) The maximum height $H$ of a projectile is given by the formula $H = \frac{u^2 \sin^2 \alpha}{2g}$,where $u$ is the initial speed and $\alpha$ is the angle of projection with the horizontal.For the first body,the angle with the horizontal is $\alpha_1 = 	heta$. Thus,$H_1 = \frac{u^2 \sin^2 	heta}{2g}$.For the second body,the angle with the vertical is $	heta$,which means the angle with the horizontal is $\alpha_2 = 90^\circ - 	heta$. Thus,$H_2 = \frac{u^2 \sin^2(90^\circ - 	heta)}{2g} = \frac{u^2 \cos^2 	heta}{2g}$.The ratio of the maximum heights is $\frac{H_1}{H_2} = \frac{u^2 \sin^2 	heta / 2g}{u^2 \cos^2 	heta / 2g} = \frac{\sin^2 	heta}{\cos^2 	heta} = 	an^2 	heta$.Therefore,the ratio is $	an^2 	heta : 1$.

$A$ body is projected at an angle $\theta$ with the horizontal. Another body is projected with the same speed at an angle $\theta$ with the vertical. The ratio of their maximum heights is:

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