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(C) In horizontal projectile motion,the horizontal component of velocity $(v_x)$ remains constant throughout the flight.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 flight.Given,initial horizontal velocity $v_x = 18 \, m s^{-1}$.Let $v_y$ be the vertical component of velocity when the body strikes the ground.The angle of impact $	heta$ with the horizontal is given by $	an 	heta = \frac{v_y}{v_x}$.Given $	heta = 45^{\circ}$,so $	an 45^{\circ} = \frac{v_y}{18}$.Since $	an 45^{\circ} = 1$,we have $1 = \frac{v_y}{18}$.Therefore,$v_y = 18 \, m s^{-1}$.

$A$ body is projected horizontally from the top of a tower with an initial velocity of $18 \, m s^{-1}$. It hits the ground at an angle of $45^{\circ}$. What is the vertical component of velocity when it strikes the ground? $(ms^{-1})$

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