Adrowsy cat spots a flowerpot that sails first up and then down past an open window. the pot was in view for a total of 0.49 s, and the top-to-bottom height of the window is 1.80 m. how high above the window top did the flowerpot go?

For this case, let's assume that the pot spends exactly half of its time going up, and half going down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take the bottom of the window to be zero on a vertical axis pointing upward. All calculations will be made in reference to this coordinate system. 

An initial condition has been supplied by the problem: 

s=1.80m when t=0.245s 

This means that it takes the pot 0.245 seconds to travel upward 1.8m. Knowing that the gravitational acceleration acts downward constantly at 9.81m/s^2, and based on this information we can use the formula:

s=(v)(t)+(1/2)(a)(t^2) 

to solve for v, the initial velocity of the pot as it enters the cat's view through the window. Substituting and solving (note that gravitational acceleration is negative since this is opposite our coordinate orientation): 

(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2 

v=8.549m/s 

Now we know the initial velocity of the pot right when it enters the view of the window. We know that at the apex of its flight, the pot's velocity will be v=0, and using this piece of information we can use the kinematic equation:

(v final)=(v initial)+(a)(t) 

to solve for the time it will take for the pot to reach the apex of its flight. Because (v final)=0, this equation will look like 

0=(v)+(a)(t) 

Substituting and solving for t: 

0=(8.549m/s)+(-9.81m/s^2)(t) 

t=0.8714s 

Using this information and the kinematic equation we can find the total height of the pot’s flight:

s=(v)(t)+(1/2)(a)(t^2) 

s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2

s=3.725m

This distance is measured from the bottom of the window, and so we will need to subtract 1.80m from it to find the distance from the top of the window: 

3.725m – 1.8m=1.925m

1.925m

temp rises, and expansion