The tiles came off the roof of the house and flew down. It flew through a window 1.8 meters high in 0.3 seconds.

The tiles came off the roof of the house and flew down. It flew through a window 1.8 meters high in 0.3 seconds. What is the distance between the roof and the top edge of the window.

The tile moves with a constant acceleration g = 9.8 m / s2 (we neglect the air resistance.

Let us find the speed that the tile had at the moment when it flew to the upper edge of the window, expressing it from the formula for displacement during rectilinear uniformly accelerated motion:

S = v0 * t + (g * t1 ^ 2) / 2;

v0 = (S1 – (g * t1 ^ 2) / 2) / t1 = (1.8 – (9.8 * 0.3 ^ 2) / 2) / 0.3 = 4.53 m / s;

Let’s find the time for which the tile flew to the upper edge of the window (gained the speed found earlier):

t2 = v0 / g = 4.53 / 9.8 = 0.46 s;

Find the distance from the edge of the roof to the top edge of the window:

S2 = (g * t2 ^ 2) / 2 = (9.8 * 0.46 ^ 2) / 2 = 1 m.

Answer: The distance from the edge of the roof to the top edge of the window is 1 m.