A drowsy cat spots a flower pot that sails first up and then down past an open window. The pot is in view for a total of 0.50s. And the top to bottom height of the window is 2.00m. How high above the window top does the flower pot go?
1.571556 meters
To find the height above the window top that the flower pot goes, we need to determine the time it takes for the pot to reach its highest point, also known as the time of flight.
Since the pot goes up and then down, we can assume it undergoes projectile motion. In projectile motion, the time of flight is twice the time it takes to reach the highest point.
Given that the pot is in view for a total of 0.50s, the time of flight is half of that:
Time of flight = 0.50s / 2 = 0.25s
Now, we can use the formulas of projectile motion to find the height above the window top.
The height reached by the pot is given by the formula:
Height = (initial vertical velocity * time) - (0.5 * acceleration due to gravity * time^2)
Since the pot starts with zero vertical velocity at the top of its trajectory, the first term will be zero in this case.
So, we are left with:
Height = -0.5 * acceleration due to gravity * time^2
The acceleration due to gravity is approximately 9.8 m/s^2. Plugging in the values, we get:
Height = -0.5 * 9.8 m/s^2 * (0.25s)^2
= -0.5 * 9.8 m/s^2 * 0.0625s^2
= -0.5 * 0.6125 m
= -0.30625 m
Since height cannot be negative in this context, the flower pot actually goes below the window top by 0.30625 meters.
Therefore, the height above the window top that the flower pot goes is approximately 0.30625 meters below the window top.
To determine the height above the window top that the flower pot goes, we can use the information provided. We know that the flower pot is in view for a total of 0.50 seconds and that the top-to-bottom height of the window is 2.00 meters.
We can use the concept of constant velocity motion to solve this problem. When an object is in free fall, it experiences constant acceleration due to gravity. We can determine the fall time and, from that, calculate the height using the equation of motion.
The total time the flower pot is in view is divided into two equal parts: the time it takes to go up and the time it takes to go back down. So, each part of the motion takes 0.50/2 = 0.25 seconds.
During this time, the flower pot goes up and then down, covering a total distance equal to the height of the window, which is 2.00 meters.
First, we need to find the time it takes for the flower pot to reach its highest point during the upward motion. Since this is a symmetrical motion, the time taken to reach the top is half of the total time for the upward motion, which is 0.25/2 = 0.125 seconds.
Using the equation of motion for the upward motion of an object in free fall:
h = ut + (1/2)gt^2
where h is the height, u is the initial velocity (which is zero here since the flower pot starts from rest), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Plugging in the values:
h = 0 + (1/2) * 9.8 * (0.125)^2
h = 0.765625 meters
Therefore, the flower pot goes approximately 0.765625 meters (or 76.5625 centimeters) above the top of the window.
You can simplify the problem a bit by realizing that the pot will take the same length of time going up as going down past the window's 2 meter height. Thus it takes 0.25 s to go from top to bottom. Let H be the verticaldistance from maximum height to the top of the window. Let t1 be the time it takes to go from maximum height H to the top of the window, and t2 is the time it takes to go from maximum height H to the bottom of the window,
t2 - t1 = 0.25 s
(1/2)g t1^2 = H
(1/2)g t2^2 = H + 2
(1/2)g (t1 + 0.25)^2 = H + 2
(1/2)g (t1 + 0.25)^2 - (1/2)t1^2 = 2
You now have only one unknown. Solve for t1 and with that you can solve for H.