A window washer drops a brush from a scaffold on a tall office building. What is the speed of the falling brush after 1.53 s? (Neglect drag forces.) The acceleration of gravity is 9.8 m/s2. Answer in units of m/s. How far has it fallen in that time? Answer in units of m.
9.8 * 1.53 = 14.994 (speed/velocity) 1/2 * 9.8 * 1.53^2 = 1/2 * 9.8 * 2.3409 = 11.47041 (distance) If 11.47041 is not the answer, then what is? The proof is in the math. I have done it following the examples of two web sites, and the same answer pops up. Maybe they want you to round up.
a) To find the speed of the brush, the equation to use is v(t) = at + v(0), where a is the acceleration of gravity; t is the elapsed time since the start of the fall, and v(0) is the initial velocity of the ball. Then we get this after 1.53 seconds: v(t) = at + v(0) v(t) = at + 0 v(t) = (9.8 m/s?)(1.53 s) v(t) = (9.8 m)(1.53 s/s?) v(t) = 14.994 m/s v(t) ≈ 15 m/s. The brush is falling at approximately 15 m/s after 1.53 seconds. b) To find the distance the brush has fallen, we use this equation: s(t) = ? at? + v(0)t + s(0). If we consider the top of the building 0, and the brush's initial velocity v(0) = 0, then the last two terms drop out of the equation above, leaving us with this: s(t) = ? at?. Plugging in a = 9.8 m/s? and t = 1.53 s, we get this: s(t) = (4.9 m/s?)(1.53 s)? s(t) ≈ 11.47 m, or 11. 5 meters rounded to the nearest tenth of a meter. The distance the brush has fallen after 1.53 seconds is approximately 11.5 meters.
