Question

A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes the pot 0.400 s to pass from the top to the bottom of this window, which is 2.10 m high.

Answer 1

Answer:

0.55 m

Explanation:

Parameters given:

Time taken to pass from the top of the window to the bottom = 0.4 s

Height of the window = 2.10 m

Given that we know the time taken for the flowerpot to fall the height of the window, we can find the initial velocity of the pot. Using one of Newton's equations of motion:

S = u*t + 0.5*g*(t^2)

Given that

g = +9.8m/s^2,

S = 2.1 m

t = 0.4 s

=> 2.1 = 0 4*u + (0.5 * 9.8 * 0.4 * 0.4)

2.1 = 0.4u + 0.784

=> 0.4u = 1.316

u = 3.29 m/s

We know that the flowerpot fell from a window above, hence, the final velocity of its fall from above is equal to the initial velocity of when it starts falling from top of the window below.

Using another equation of motion, we have that:

v^2 = u^2 + 2gs

Where

s = height of the window above to the top of the window below

v = 3.29 m/s

u = 0 m/s

3.29^2 = 0 + 2*9.8*s

10.82 = 19.6s

=> s = 0.55 m

Hence, the top of the window above is 0.55m far from the top of the window below.