Problem 46.

A flowerpot falls from a windowsill 25.0 m above the sidewalk.      

(a) how fast is the flowerpot moving when it strikes the ground?

(b) how much time does a passerby on the sidewalk below have to move out of the
        way before the flowerpot hits the ground?

Solution:

This is the free fall motion with constant acceleration. The equations which describe this motion are the following:

where we introduced an axis y with upward direction. The origin of axis y is the initial position of the flowerpot. The initial velocity of flowerpot is 0.

We know the final position of the flowerpot: the coordinate of the final point is -25 m.

(a) We substitute this coordinate in the third equation an obtain the velocity:

The negative sign means that the direction of velocity is downward. The magnitude of velocity is 22 m/s.

(b) The traveled time can be found from the first equation in the above system. We substitute y(t)=-25 and obtain time: