Question

A window washer drops a brush from a scaffold
on a tall office building.
What is the speed of the falling brush after
3.08 s? (Neglect drag forces.) The acceleration
due to gravity is 9.8 m/s^2.
Answer in units of m/s.

Answer 1

Answer on Question 62464, Physics, Solid State Physics

Question:

A window washer drops a brush from a scaffold on a tall office building. What is the speed of the falling brush after $3.08\,s$? (Neglect drag forces.) The acceleration due to gravity is $9.8\,\frac{m}{s^2}$. Answer in units of $\frac{m}{s}$.

Solution:

We can find the speed of the falling brush after $3.08\,s$ from the kinematic equation:

$$v = v_0 + at,$$

here, $v_0$ is the initial speed of the falling brush, $t$ is the time, $a = g = 9.8\,\frac{m}{s^2}$ is the acceleration due to gravity (we take the downwards to be the positive direction, thus the acceleration due to gravity will be positive).

Since initially the brush was at rest, $v_0 = 0\,\frac{m}{s}$, and we get:

$$v(3.08\,s) = gt = 9.8\,\frac{m}{s^2} \cdot 3.08\,s = 30.18\,\frac{m}{s}.$$

Answer:

$$v(3.08\,s) = 30.18\,\frac{m}{s}.$$

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