Question

It takes 0.13s for a dropped object to pass a window that is 1.40m tall. From what height above the top of the window was the object released?

Answer 1

Step-by-step explanation:

d = 1.4 t = .13

df = do + vo t + 1/2 a t^2 <=====sub in the values

1.4 = 0 + vo (.13) + 1/2 (9.81)(.13^2) shows vo, the velocity at the top of the window to be vo = 10.13158 m/s

Now using vo as vf when the object reaches the top of the window:

vf = vo + at ( remember vo = 0 as it is released )

10.13158 = 0 + 9.81 t

shows time to top of window to be t = 1.03278seconds

then d = 1/2 a t^2 shows

d = 5.23 m above the top of the window release point