A ball is dropped from the top of a building. The ball takes `0.50`s to fall past the 3m length of a window, which is some distance below the top of the building.
(a) How fast was the ball going as it passed the top of the window?
(b) How far is the top of the window from the point at which the ball was dropped?
Assume acceleration g in free fall due to gravity to be `10 m//s^(2)` downwards.

Explanation

The ball is dropped, so it start falling the top of the building with zero initial velocity `(v_(o)=0)`. The motion diagram is shown with the given information in the adjoining figure.
Using the first equation of the constant acceleration motion, we have
`v_(t)=v_(o)+at rarr v=0+10t=10t` ...(i)
`v'=0+10(t+0.5)=10t+5` ...(ii)
Using values of v and v' in following equation, we have
`x-x_(o)=((v_(o)+v)/2)t rarr` window height `((v+v')/2)xx0.5rArr t=0.35s`
(a) From equation (i), we have `v=10t=3.5 m//s`
(b) From following equation, we have `" " x-x_(o)=((v_(o)+v)/2)t rarr h=((0+v)/2)t=61.25` cm