A stone falls vertically past a window 2m high in 0.2s .find the height above the top of the window from which the stone was dropped
To find the height above the top of the window from which the stone was dropped, we can use the equations of motion.
The equation that relates displacement, initial velocity, time, and acceleration for an object in free fall is:
s = ut + (1/2)at^2
Where:
s is the displacement (height),
u is the initial velocity (which is 0 since the stone was dropped),
t is the time, and
a is the acceleration due to gravity (9.8 m/s^2).
In this case, we want to find the height above the top of the window, so we need to calculate the total displacement of the stone.
Given:
Window height (h) = 2m
Time taken (t) = 0.2s
Acceleration due to gravity (a) = 9.8 m/s^2
Since the stone is falling vertically, the displacement can be considered as negative. Therefore, the equation becomes:
-2 = 0 - (1/2)(9.8)(0.2)^2
Simplifying the equation:
-2 = -0.98 * 0.04
-2 = -0.0392
Therefore, the height above the top of the window from which the stone was dropped is approximately 0.0392 meters (or 3.92 centimeters).
the distance fallen in t seconds is 1/2 gt^2 ≈ 5t^2
If it reaches the top of the window at time t, then we have
5(t + 0.2)^2 - 5t^2 = 2.0
t^2 + .4t + .04 - t^2 = .4
.4t = .36
t = 0.9
So, how far did it fall during the first 0.9 seconds?
You can use g = 9.8 instead of 10 if you want.