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(A) The time of flight $T$ for a projectile is given by $T = \frac{2 u \sin 	heta}{g}$,where $u$ is the initial velocity and $	heta$ is the angle of

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{g T^2}{2 R}ight)$

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(A) The time of flight $T$ for a projectile is given by $T = \frac{2 u \sin 	heta}{g}$,where $u$ is the initial velocity and $	heta$ is the angle of projection.From this,we can express the initial velocity as $u = \frac{g T}{2 \sin 	heta}$.The horizontal range $R$ is given by $R = \frac{u^2 \sin(2 	heta)}{g} = \frac{2 u^2 \sin 	heta \cos 	heta}{g}$.Substituting $u$ into the range formula: $R = \frac{2}{g} \left(\frac{g T}{2 \sin 	heta}ight)^2 \sin 	heta \cos 	heta$.Simplifying the expression: $R = \frac{2}{g} \cdot \frac{g^2 T^2}{4 \sin^2 	heta} \cdot \sin 	heta \cos 	heta$.$R = \frac{g T^2}{2} \cdot \frac{\cos 	heta}{\sin 	heta} = \frac{g T^2}{2 	an 	heta}$.Rearranging for $	an 	heta$: $	an 	heta = \frac{g T^2}{2 R}$.Therefore,the angle of projection is $	heta = 	an^{-1}\left(\frac{g T^2}{2 R}ight)$.

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

If the time of flight of a bullet over a horizontal range $R$ is $T$,then the angle of projection with the horizontal is ......

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