An object falling in a vertical line passes a window 3 m high in 1/6 seconds. Find the distance above the top of the window from which the object let fall.
No idea about this question at all.
Not sure if I'm right, we'll have to see what Trebla says, but I think the answer is 15m.
s=ut + (1/2)at^2 to find the u, the initial velocity at the start of the window, s=3, t=1/6, a=9.8. So u=17.18ms^-1.
Then use v=u+at to find the final velocity (at the bottom of the 3m window). (u=17.18, t=1/6, a=9.8, therefore v=18.8ms^-1).
Then use v^2 = u^2 + 2as to find the whole distance the object fell, so 18.8^2 = 0^2 + 2(9.8)s
s=18.03, minus 3 to find the distance before the window.
Distance = 15m
