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(C) In horizontal projectile motion,the horizontal component of velocity $(v_x)$ remains constant throughout the motion.Given,initial horizontal veloc

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$18$

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(C) In horizontal projectile motion,the horizontal component of velocity $(v_x)$ remains constant throughout the motion.Given,initial horizontal velocity $u_x = 18 \, ms^{-1}$.Therefore,at the time of striking the ground,$v_x = 18 \, ms^{-1}$.The angle of impact with the ground is given as $	heta = 45^o$.The relationship between the velocity components and the angle is given by $	an 	heta = \frac{v_y}{v_x}$.Substituting the values: $	an 45^o = \frac{v_y}{18}$.Since $	an 45^o = 1$,we have $1 = \frac{v_y}{18}$.Thus,$v_y = 18 \, ms^{-1}$.

$A$ body is projected horizontally from a tower with a velocity of $18 \, ms^{-1}$. It strikes the ground at an angle of $45^o$. The vertical component of the velocity when it strikes the ground is ........ $ms^{-1}$.

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