A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of 0.37 s, and the top-to-bottom height of the window is 2.20 m. How high above the window top did the flowerpot go?
S= ut +(1/2)at^2
S= (0)(0.215) + (0.5)(9.8)(0.215)^2
S= 0.2265025m
That's what I did and the site tells me I'm wrong. Please help.
Duplicate post--already answered.
To answer this question, you can use the kinematic equation for displacement:
S = ut + (1/2)at^2
Where:
S is the displacement (height) of the flowerpot,
u is the initial velocity of the flowerpot,
a is the acceleration due to gravity (approximately 9.8 m/s^2),
and t is the time for which the flowerpot was in view (0.37 s).
However, in this case, you don't know the initial velocity (u) of the flowerpot, but you do know that it starts from rest (drowsy cat spots the flowerpot). This means that the initial velocity (u) can be assumed to be zero.
So, the equation simplifies to:
S = (1/2)at^2
Substituting the known values:
S = (1/2)(9.8)(0.37)^2
S = (4.9)(0.1369)
S ≈ 0.6692 m
Therefore, the flowerpot went approximately 0.6692 meters above the top of the window.