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A pen rolls off a 0.55–meter high table with an initial horizontal velocity of 1.2 meters/second. At what horizontal distance from the base of the table will the pen land?
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h=1/2*gt^2 t=sqrt(2*h/g) let s is the horozontal distence s=(horizontal velocity given)*t =1.2*sqrt(2*0.55/9.8)=0.4 m hope u'll understand
Yes, procedure for working problems like this is to find how long the object is in motion (by determining the time for it to fall to the ground), then multiply the time by the horizontal component of the velocity. here are the same equations, possibly a bit easier to read: \[h = \frac{1}{2}gt^2\]\[t=\sqrt{\frac{2h}{g}}\]\[x = v_h t\]\[v_h=1.2\text{ m/s}\]\[g=9.8\text{ m/s}^2\]
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