Question

A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.64 s, and the top-to-bottom height of the window is 1.66 m. How high above the window top does the flowerpot go? (Assume that in both sailing upward and then falling downward, the flowerpot is seen to span the full vertical height of the window.)

Answer 1

Here the time for going up will be 0.64/2 = 0.32 s as time in going up and coming down are equal,

Let the velocity at window top be v, then velocity at window bottom u = v - at

= v + gt

= v+9.8*0.32

= v + 3.136

Now using third equation of motion,

v^2 = u^2 + 2as where s is distance, a is acceleration

v^2 = (v + 3.136)^2 + 2*-9.8*1.66

v = 3.6195 m/s

Distance it rises above window top = v^2/2g

= 3.6195^2/[2*9.8]

= 0.6684 m answer