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(C) The horizontal component of the initial velocity is $u_x = u \cos 	heta$.Since there is no acceleration in the horizontal direction,the horizonta

Vedclass · Accepted Answer

$\frac{u^2 \sin 2	heta}{g}$

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(C) The horizontal component of the initial velocity is $u_x = u \cos 	heta$.Since there is no acceleration in the horizontal direction,the horizontal velocity remains constant throughout the motion.The time of flight $T$ for a projectile is given by $T = \frac{2u \sin 	heta}{g}$.The horizontal range $R$ is the product of the horizontal velocity and the time of flight:$R = u_x 	imes T$$R = (u \cos 	heta) 	imes \left( \frac{2u \sin 	heta}{g} ight)$Using the trigonometric identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we get:$R = \frac{u^2 (2 \sin 	heta \cos 	heta)}{g} = \frac{u^2 \sin 2	heta}{g}$.

When a body is thrown with a velocity $u$ making an angle $\theta$ with the horizontal plane,the maximum distance covered by it in the horizontal direction is:

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