The maximum horizontal range of a projectile | vedclass

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Question

$\mathrm{R}=\frac{\mathrm{u}^{2} \sin 2 	heta}{\mathrm{g}}$ so that $\mathrm{R}_{\max }=\frac{\mathrm{u}^{2}}{\mathrm{g}}=160 \mathrm{m}$

$\mathrm

Vedclass · Accepted Answer

$20$

Vedclass · Answer

$\mathrm{R}=\frac{\mathrm{u}^{2} \sin 2 	heta}{\mathrm{g}}$ so that $\mathrm{R}_{\max }=\frac{\mathrm{u}^{2}}{\mathrm{g}}=160 \mathrm{m}$

$\mathrm{H}_{\max }=\frac{\mathrm{u}^{2} \sin ^{2} 	heta}{2 \mathrm{g}}$

$=\frac{1}{2} 	imes 160 	imes \sin ^{2} 30^{\circ}=20$

The maximum horizontal range of a projectile is $160\, m$. When the projectile is thrown with the same speed at an elevation of $30^o$ from the horizontal, it will reach to the maximum height of ......... $m$

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