A window washer drops a brush from a scaffold on a tall office building.

What is the speed of the falling brush after 2.48 s? (Neglect drag forces.) The acceleration due to gravity is 9.8 m/s 2.

Answer in units of m/s

To find the speed of the falling brush after 2.48 s, we can use the equation of motion:

v = u + at

where:
v is the final velocity (speed) of the brush,
u is the initial velocity (which is zero since the brush was dropped),
a is the acceleration due to gravity (9.8 m/s^2),
and t is the time (2.48 s).

Substituting the given values into the equation, we get:

v = 0 + (9.8 m/s^2) * (2.48 s)

Now, we can solve this equation to find the speed of the falling brush:

v = 9.8 m/s^2 * 2.48 s
v = 24.304 m/s

Therefore, the speed of the falling brush after 2.48 s is approximately 24.304 m/s.