The maximum horizontal range of a projectile | vedclass

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Question

(A) The maximum horizontal range is given by $R_{\max} = \frac{u^2}{g} = 400\;m$.The maximum height attained by a projectile is given by $H = \frac{u^

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) The maximum horizontal range is given by $R_{\max} = \frac{u^2}{g} = 400\;m$.The maximum height attained by a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.For maximum horizontal range,the angle of projection is $	heta = 45^\circ$.Substituting $	heta = 45^\circ$ into the height formula:$H = \frac{u^2 \sin^2 45^\circ}{2g} = \frac{u^2 (1/\sqrt{2})^2}{2g} = \frac{u^2}{2g 	imes 2} = \frac{u^2}{4g}$.Since $R_{\max} = \frac{u^2}{g} = 400\;m$,we have:$H = \frac{1}{4} 	imes R_{\max} = \frac{1}{4} 	imes 400\;m = 100\;m$.

The maximum horizontal range of a projectile is $400\;m$. What is the maximum height attained by it (in $;m$)?

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