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Question

(A) Let the initial velocity be $u = 30 \ m \ s^{-1}$ and the angle of projection be $	heta$. The maximum height reached by the projectile is $H = \f

Vedclass · Accepted Answer

$60 \sqrt{2} \ m$

Vedclass · Answer

(A) Let the initial velocity be $u = 30 \ m \ s^{-1}$ and the angle of projection be $	heta$. The maximum height reached by the projectile is $H = \frac{u^2 \sin^2 	heta}{2g} = 30 \ m$.Substituting the values: $\frac{(30)^2 \sin^2 	heta}{2 	imes 10} = 30 \implies \frac{900 \sin^2 	heta}{20} = 30 \implies 45 \sin^2 	heta = 30 \implies \sin^2 	heta = \frac{30}{45} = \frac{2}{3}$.Thus,$\sin 	heta = \sqrt{\frac{2}{3}}$ and $\cos 	heta = \sqrt{1 - \frac{2}{3}} = \frac{1}{\sqrt{3}}$.The range of the projectile is $R = \frac{u^2 \sin 2	heta}{g} = \frac{u^2 (2 \sin 	heta \cos 	heta)}{g}$.$R = \frac{30^2 	imes 2 	imes \sqrt{\frac{2}{3}} 	imes \frac{1}{\sqrt{3}}}{10} = \frac{900 	imes 2 	imes \frac{\sqrt{2}}{3}}{10} = 90 	imes 2 	imes \frac{\sqrt{2}}{3} = 60 \sqrt{2} \ m$.

$A$. Velocity of the body after $1 \,s$ (in $ms^{-1}$)	$I$. $5$
$B$. Horizontal displacement of the body after $1 \,s$ (in $m$)	$II$. $20$
$C$. Vertical displacement of the body after $1 \,s$ (in $m$)	$III$. $10$
$D$. Vertical velocity of the body after $1 \,s$ (in $ms^{-1}$)	$IV$. $22.4$

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