A falling stone takes 0.300 s to travel past a window 2.40 m tall. From what height above the top of the window did the stone fall?

Question

A falling stone takes 0.300 s to travel past a window 2.40 m tall. From what height above the top of the window did the stone fall?

acceleration (a) =

Vi =

Vf =

∆d =

∆t=

Solve...

Answer 1

Let this hite is h. Then h = gt²/2. From the other side h + Δh = g(t + Δt)²/2,

where g = 9.80 m/sec² is accelaration of gravity, h is unknown height above window, Δh is height of window,

Δt = 0.300 sec.

From second equation h + Δh = gt²/2 + gt·Δt + g(Δt)²/2 or Δh = gt·Δt + g(Δt)²/2;

From here t = Δh/(g·Δt) - Δt/2 = 2.40m/(9.80m/sec²· 0.300sec) - 0.300sec/2 = 0.66633 sec.

Now we can find h = 9.80 m/sec²·0.66633²sec²/2 = 2.176 m.

Answer: stone falled from 2.176 m above window