A window washer drops a brush from a scaffold on a tall office building.
What is the speed of the falling brush after
2.03 s? (Neglect drag forces.) The acceleration due to gravity is 9.8 m/s
2
.
Answer in units of m/s
i got 20.19 but its wrong
velocity = v = -gt = -9.8*2.03 = -19.89
so speed = 19.9 m/s
To solve this problem, you need to use the equations of motion and the acceleration due to gravity. The equation you will use is:
v = u + at
where:
v is the final velocity,
u is the initial velocity (which is 0, as the brush is initially at rest),
a is the acceleration (which is equal to the acceleration due to gravity, 9.8 m/s^2),
t is the time (2.03 s in this case).
Substituting the values into the equation, we have:
v = 0 + (9.8 m/s^2) * (2.03 s)
v = 0 + 19.894 m/s
v ≈ 19.9 m/s
Therefore, the speed of the falling brush after 2.03 s is approximately 19.9 m/s.