Vikas TU
Last Activity: 7 Years ago
Let V0 = the underlying speed of the stone as it crossed the top edge of the window
Let h = the tallness over the window that the stone was dropped with no underlying speed.
Let g = the speeding up because of gravity = 9.8 m/s²
Let t = the time that it takes to fall past the window = 0.2 s
Let d = the tallness of the window = 1 m
Utilizing the separation condition for steady speeding up with an underlying speed:
d = (V0)(t) + (1/2)(g)(t²)
The underlying speed condition is:
V0 = √{(2)(g)(h)}
Substitute √{(2)(g)(h)} for V0:
d = √{(2)(g)(h)}t + (1/2)(g)(t²)
Unravel for h:
√{(2)(g)(h)}t = d - (1/2)(g)(t²)
(2)(g)(h)t² = {d - (1/2)(g)(t²)}²
h = {d - (1/2)(g)(t²)}²/{2(g)(t²)}
h ≈ 0.824 m
