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(B) The motion of the ball is a horizontal projectile motion.For vertical motion, the initial vertical velocity $u_y = 0 	ext{ m/s}$.The vertical dis

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$2 	ext{ s}$

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(B) The motion of the ball is a horizontal projectile motion.For vertical motion, the initial vertical velocity $u_y = 0 	ext{ m/s}$.The vertical displacement is $H = 19.6 	ext{ m}$.The acceleration due to gravity is $g = 9.8 	ext{ m/s}^2$.Using the equation of motion $H = u_y t + \frac{1}{2} g t^2$, we get:$19.6 = 0 	imes t + \frac{1}{2} 	imes 9.8 	imes t^2$$19.6 = 4.9 	imes t^2$$t^2 = \frac{19.6}{4.9} = 4$$t = \sqrt{4} = 2 	ext{ s}$.Thus, the ball will take $2 	ext{ s}$ to hit the ground.

$A$ ball is projected horizontally with a velocity of $5 \text{ m/s}$ from the top of a building $19.6 \text{ m}$ high. How long will the ball take to hit the ground?

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