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 seconds, we can use the kinematic equation:
v = u + at
where:
v is the final velocity (speed) of the brush after 2.48 seconds,
u is the initial velocity (which in this case is zero because the brush was dropped),
a is the acceleration due to gravity (which is -9.8 m/s^2 because it acts downward),
and t is the time elapsed (2.48 seconds).
Plugging in the values into the equation, we have:
v = 0 + (-9.8) * 2.48
Simplifying the equation, we get:
v = -24.304 m/s
However, since speed is a scalar quantity (magnitude only), we take the absolute value of the answer:
v = 24.304 m/s
Therefore, the speed of the falling brush after 2.48 seconds is 24.304 m/s.