A falling stone take 0.2s to fall post a window with. Hight 1m from how far above the top of window stone dropped
To find out how far above the top of the window the stone was dropped, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance traveled (1m above the window)
u = initial velocity (0 m/s since the stone is dropped)
t = time taken (0.2s)
a = acceleration due to gravity (-9.8 m/s^2)
Plugging in the values into the equation, we have:
1m = 0(0.2s) + (1/2)(-9.8m/s^2)(0.2s)^2
Simplifying the equation, we get:
1m = (1/2)(-9.8m/s^2)(0.04s^2)
1m = -0.98m/s^2 * 0.04s^2
1m = -0.0392m
Since distance cannot be negative, we can ignore the negative sign. Therefore, the stone was dropped from a height of approximately 0.0392 meters above the top of the window.
avg vel is 5 m/s (Va)
during .2 s passage, accel (A) is ... .2 * g
find vel at top of window .. Vt = Va - (A / 2)
find fall time (Tf) to top of window ... Vt / g
drop distance is ... 1*2 * g * Tf^2