a stone drops from the edge of the roof. it passes a window 2m high in 0.1s.how far is the roof from the windo?
The stone drops ... so v(i) = 0 m/s Take down as the positive direction ... so a = g = 9.8 m/s? Let the distance from the roof to the top of the window = y v(f)? = v(i)? + 2as v(f)? = 0 + 19.6y = 19.6y so the velocity of the stone when it reaches the top of the window = √(19.6y) Now consider the stones motion as it drops past the window: v(i) = √(19.6y) ... [the speed at the top of the window] v(f) = v(i) + at so v(f) = √(19.6y) + 9.8 * 0.1 v(f) = √(19.6y) + 0.98 ... [and that's the stone's velocity at the bottom of the window] also v(f)? = v(i)? + 2as so [√(19.6y) + 0.98]? = [√(19.6y)]? + 2 * 9.8 * 2 19.6y + 1.96√(19.6y) + 0.98? = 19.6y + 39.2 1.96√(19.6y) = 39.2 - 0.98? √(19.6y) = [39.2 - 0.98?] / 1.96 √(19.6y) = 19.51 19.6y = 19.51? y = 19.51? / 19.6 y = 19.42 so the roof is 19.42 m from the window
