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(C) The maximum height of a projectile is given by $H = \frac{v^2 \sin^2 	heta}{2g}$.We are looking for the vertical component of velocity $v_y$ at a

Vedclass · Accepted Answer

$\frac{v \sin 	heta}{\sqrt{2}}$

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(C) The maximum height of a projectile is given by $H = \frac{v^2 \sin^2 	heta}{2g}$.We are looking for the vertical component of velocity $v_y$ at a height $h = \frac{H}{2} = \frac{v^2 \sin^2 	heta}{4g}$.Using the kinematic equation $v_y^2 = u_y^2 - 2gh$,where $u_y = v \sin 	heta$ is the initial vertical velocity:$v_y^2 = (v \sin 	heta)^2 - 2g \left( \frac{v^2 \sin^2 	heta}{4g} ight)$$v_y^2 = v^2 \sin^2 	heta - \frac{v^2 \sin^2 	heta}{2}$$v_y^2 = \frac{v^2 \sin^2 	heta}{2}$Taking the square root of both sides,we get:$v_y = \frac{v \sin 	heta}{\sqrt{2}}$Thus,the correct option is $(C)$.

$A$. 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 $ms^{-1}$)	$IV$. $22.4$

$A$ projectile is thrown with velocity $v$ at an angle $\theta$ with the horizontal. When the projectile is at a height equal to half of the maximum height,the vertical component of the velocity of the projectile is ...........

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