Video Example 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 6 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 6 s is that we are dealing with a single instant of time (t = 6), 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 = 6 to t = 6.1: average velocity = change in position time elapsed = s(6.1) − s(6) 0.1

58.8m/s

Step-by-step explanation:

Using the equation of motion of a freely falling body:

v = u + gt(1)

u = 0, g= +9.8 m/s²(g is a constant called acceleration due to gravity), t = 6s

v = 0 + 9.8× 6

v = 58.8 m/s

Using Galileo's model:

s = 4.9t²(1)

The velocity is obtained by finding the differential of s with respect to t

ds/dt = 2 × 4.9 t(2)

At time t= 6, the model support the velocity as calculated (1) above at a particular time t=6:

ds/dt = 2 ×4.9(6) = 58.8m/s

The velocity 58.8 gives the

Using the model, we can calculate the time required for the ball to travel just before hitting the ground,

450 = 4.9t²

t² = 450/4.9

t = √(450/4.9)

t = 9.58s

Using (1) v = gt = 9.8 × 9.58

v = 93.88

= 93.9m/s(This is the velocity before hitting the ground).

Pls, note the velocity 58.8m/s is the velocity at a particular time in the journey, since body is under a constant acceleration the velocity changes with time as given by the velocity 93.8m/s.

Hence the velocity change

between t1 = 6.0 and t2 = 6.1,

ds/dt = ∆v

= 2 × 4.9∆t

= 2 × 4.9 (6.1 - 6.0)

= 9.8 ×0.1

= 0.98m/s.

2^y = ln 5, or   y(ln 2) = ln 5.   solving for y,   y = (ln 5) / (ln 2).   y=5

step-by-step explanation:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.