A man runs across the roof of a tall building | vedclass

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Question

(A) Given:Horizontal speed of the man,$u_x = 9 \, m/s$Horizontal distance between the two buildings,$x = 10 \, m$Vertical height difference,$h = 9 \,

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(A) Given:Horizontal speed of the man,$u_x = 9 \, m/s$Horizontal distance between the two buildings,$x = 10 \, m$Vertical height difference,$h = 9 \, m$Acceleration due to gravity,$g = 10 \, m/s^2$Let $t$ be the time taken to cover the vertical distance $h$. Using the equation of motion in the vertical direction:$h = u_y t + \frac{1}{2} g t^2$Since the man jumps horizontally,the initial vertical velocity $u_y = 0$.$9 = 0 	imes t + \frac{1}{2} 	imes 10 	imes t^2$$9 = 5 t^2$$t^2 = \frac{9}{5} = 1.8$$t = \sqrt{1.8} \approx 1.34 \, s$Now,calculate the horizontal distance $R$ covered by the man in this time:$R = u_x 	imes t = 9 	imes \sqrt{1.8} = 9 	imes 1.3416 \approx 12.07 \, m$Since the horizontal distance covered $(12.07 \, m)$ is greater than the distance between the buildings $(10 \, m)$,the man will be able to land on the next building.

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