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7. -S POINTS SCALC8 1.4.AE.003. MY NOTES ASK YOUR TEACHER Interval EXAMPLE 3 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 7 seconds. Average velocity (m/s) 73.5 69.09 75705 $ 7.01 68.845 68 649 68 6049 157.001 SOLUTION Through experiments carried out four centuries ago, Galleo discovered that the distance fallen by any freely falling body is proportional to the square of the time it has been failing. (This model for free fall neglects alr resistance.) If the distance fallen after t seconds is denoted by s(t) and measured in meters, then Galileo's www is expressed by the equation (0) 4.9 The difficulty in finding the velocity after 7 is that we are dealing with a single instant of time it 71, so no time interval is involved. However, we can approximate the desired quantity by computing the average velocky over the brie time interval of a tenth of a second from 7 tot 7.1: change in position average velocity - time lapsed Video Example m . 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 ms rounded to one decimal pbce). The instantaneous velocity when - is defined to be the limiting value of these average veces over shorter and shorter time periods that start at 7. Thus the instantaneous) velocity after 7 is the following (Round your answer to one decimal place) Need Help? Read Tate Tutor

Answer 1

Since ball down with are freely going the effect of gravity By Newton's 2nd law - 450m Distance travelled by ball round Sitla utt & gt² here u zo & initial velocity Sita **9.943 Setle 4.942 – 0 The velocity of ball after 7 second] w2a 4² tags = v²= 2g x 4.9 +2 for t= 7 see { ideal velocity. y I It is very difficut to v2= 2x 9.8 84 +9 x 72 measured ge2= 2 x 4.9X4.9 x 72 V = 8x4987 V = 62.6 mis

Since average velocity given by Sherge in position - time periodo For at Lrt Vany = $(7:1) – S(1) 71-70 Vary 4.9 (7.112 _ 4.9 (7.012 Loit - Vang = 4.9 x 24.2.X 0% Vang = 6g.og mes] interval qct <q.05 for Vous = $17.05) – 51710) 0.05 - Vonya. 419((7.05)2–07.092) - 0:05 vaus = 4.9 x 14.05X0.05 oros 60 845 mis! vary =

fori acts 7.0% Vany 5672012-54702 oroko Vary = 4.9667.01) 2 (7.022) 0.0L ---- Vary - 4.9% 14:01 Xoret Det Vany s. 68.649 mis for at s9.04 Vony = 5176001 - 51710) oroot Vary 4.9 (7.001) ² (7-072) oroot Vany - 4.9 x 14.001x0100) : icolour vary - 68.6049 mis/ Hence, average velocity for a Sharten period is closer to 68.6049 ms While instantaneous velocity at t=1 is 60 bmis