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(B) The initial velocity is given by $\vec{u} = u_0 \hat{i} + v_0 \hat{j}$.Here,the horizontal component of velocity is $u_x = u_0$ and the vertical c

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$\frac{2u_0v_0}{g}$

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(B) The initial velocity is given by $\vec{u} = u_0 \hat{i} + v_0 \hat{j}$.Here,the horizontal component of velocity is $u_x = u_0$ and the vertical component is $u_y = v_0$.The acceleration in the $x$-direction is $a_x = 0$ and in the $y$-direction is $a_y = -g$.The time of flight $T$ is the time taken for the vertical displacement to become zero: $y = u_y T - \frac{1}{2} g T^2 = 0$.$v_0 T - \frac{1}{2} g T^2 = 0 \implies T = \frac{2v_0}{g}$.The maximum displacement in the $x$-direction (horizontal range $R$) is given by $R = u_x 	imes T$.Substituting the values,$R = u_0 	imes \left( \frac{2v_0}{g} ight) = \frac{2u_0v_0}{g}$.

$A$ projectile is projected from the ground with an initial velocity $\vec{u} = u_0 \hat{i} + v_0 \hat{j}$. If the acceleration due to gravity $g$ is along the negative $y$-direction,find the maximum displacement in the $x$-direction (horizontal range).

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