Question

A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.50 s, and the top-to-bottom height of the window is 2.00 m. How high above the window top does the flowerpot go?

Answer 1

Answer:

2.34 m

Explanation:

The cat saw the pot for 0.25 s each time. use the second equation of motion to find the velocity of the pot.

x= ut + 0.5 at²

x = 2.00 m

t = 0.25 s

a = - g = -9.8 m/s²

2.00 -0 = u (0.25) - 0.5 (9.8) (0.25)²

⇒u = 9.225 m/s

Use the third equation of motion:

2as = v²-u²

at highest point, final velocity is zero. v = 0.  a =-g = -9.8 m/s²

s = -(9.225)²/(-9.8×2) =4.34 m

The height above the window top:

4.34 m - 2 m = 2.34 m