Question

Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 3 seconds. SOLUTION Through experiments carried out four centuries ago, Galileo discovered that the distance fallen by any freely falling body is proportional to the square of the time it has been falling. (This model for free fall neglects air resistance.) If the distance fallen after t seconds is denoted by s(t) and measured in meters, then Galileo's law is expressed by the equation s(t) = 4.9t2. The difficulty in finding the velocity after 3 s is that we are dealing with a single instant of time (t = 3), so no time interval is involved. However, we can approximate the desired quantity by computing the average velocity over the brief time interval of a tenth of a second from t = 3 to t = 3.1: average velocity = change in position time elapsed = s(3.1) − s(3) 0.1 = 4.9 2 − 4.9 2 0.1 = m/s. The table shows the results of similar calculations of the average velocity over successively smaller time periods. It appears that as we shorten the time period, the average velocity is becoming closer to m/s (rounded to one decimal place). The instantaneous velocity when t = 3 is defined to be the limiting value of these average velocities over shorter and shorter time periods that start at t = 3. Thus the (instantaneous) velocity after 3 s is the following. (Round your answer to one decimal place.) v = m/s

Answer 1

Answer:

1350

Step-by-step explanation:

460 X 3