This is the problem I am working on and for some reason, I cannot get it... In a

Question

This is the problem I am working on and for some reason, I cannot get it... In a mall, a shopper rides up an escalator between floors. Atthe top of the escalator, the shopper turns right and walks 15.0 mto a store. The magnitude of the shopper's displacement from thebottom of the escalator is 16.4 m. The vertical distance betweenthe floors is 4.40 m. At what angle is the escalator inclined abovethe horizontal? This is the problem I am working on and for some reason, I cannot get it... In a mall, a shopper rides up an escalator between floors. Atthe top of the escalator, the shopper turns right and walks 15.0 mto a store. The magnitude of the shopper's displacement from thebottom of the escalator is 16.4 m. The vertical distance betweenthe floors is 4.40 m. At what angle is the escalator inclined abovethe horizontal?

Explanation / Answer

a) vx = 8.5m/s since it is fly horizontally vy = gt v = [(vx)2 +(vy)2] when t=0, vy = 0, v = vx let v = 2vx speed double, we have: v=2vx = [(vx)2 +(vy)2] =>4vx2 =vx2 + vy2 => vy2 = g2 t2= 3vx2 =>t = (3)vx /g = 3*8.5m/s /9.8m/s2 = 1.25s =>t = (3)vx /g = 3*8.5m/s /9.8m/s2 = 1.25s

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