I am looking out a window that is 1.3m tall. A water balloon suddenly drops across the entire length of the window in .22 seconds. How far above the window was it dropped?
distance= vi*t+1/2 g t^2
1.3=vi (.22)+4.9 (.22^2) solve for vi, at the top of the windown.
now, distance to roof:
vabove^2=2 g h solve for h.
To find out how far above the window the water balloon was dropped, we can use the equation of motion under constant acceleration. The equation is as follows:
s = ut + (1/2)at^2
where:
s = displacement
u = initial velocity
t = time taken
a = acceleration
In this case, the water balloon is dropped, so the initial velocity (u) is 0 m/s. We need to find the displacement (s) and the time taken (t).
From the information given in the question, the height of the window is 1.3m, and the time taken for the water balloon to fall across the entire length of the window is 0.22 seconds.
Since the displacement (s) we want to find is above the window, we will consider it as positive. The acceleration due to gravity (a) is always acting downwards, so it will be negative.
Plugging the values into the equation, we have:
1.3 = 0 + (1/2)(-9.8)t^2
Simplifying the equation gives:
1.3 = (-4.9)t^2
Rearranging the equation to solve for t^2, we have:
t^2 = (1.3) / (-4.9)
t^2 ≈ -0.2653
Since time cannot be negative, we can conclude that there is an error or inconsistency in the given information. The time taken cannot be negative. Please verify the values provided.