A window washer drops a bucket from his perch on a tall building. Through her window, an accountant in an office on a lower floor sees the bucket fall. It takes 0.30 s to fall past her window that is 2.2 m tall.
How high is the window washer above the bottom of her window?
What is the velocity of the bucket when it reaches the bottom of the window?
The correct answer and explanation is :
Let’s solve this step by step:
Given data:
- Time taken to pass the window: ( t = 0.30 ) s
- Height of the window: ( h = 2.2 ) m
- Acceleration due to gravity: ( g = 9.81 ) m/s²
We need to find:
- The velocity of the bucket at the bottom of the window.
- The height from which the bucket was dropped.
Step 1: Finding the velocity at the bottom of the window
Using the kinematic equation for uniformly accelerated motion:
[
h = v_i t + \frac{1}{2} g t^2
]
where:
- ( h = 2.2 ) m
- ( t = 0.30 ) s
- ( g = 9.81 ) m/s²
- ( v_i ) is the velocity of the bucket at the top of the window.
Rearrange to solve for ( v_i ):
[
v_i = \frac{h – \frac{1}{2} g t^2}{t}
]
Substituting the values:
[
v_i = \frac{2.2 – \frac{1}{2} (9.81) (0.30)^2}{0.30}
]
[
v_i = \frac{2.2 – 0.44145}{0.30} = \frac{1.75855}{0.30} = 5.86 \text{ m/s}
]
Now, using the velocity equation:
[
v_f = v_i + g t
]
[
v_f = 5.86 + (9.81 \times 0.30)
]
[
v_f = 5.86 + 2.943 = 8.80 \text{ m/s}
]
So, the velocity of the bucket at the bottom of the window is 8.80 m/s.
Step 2: Finding the height of the window washer
The bucket starts from rest, so using:
[
v_i^2 = 2 g h’
]
[
h’ = \frac{v_i^2}{2g} = \frac{(5.86)^2}{2 (9.81)}
]
[
h’ = \frac{34.34}{19.62} = 1.75 \text{ m}
]
Thus, the window washer is 1.75 m above the top of the window, or 3.95 m above the bottom of the window.
