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.