Question

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 2 seconds.

Answer 1

225 m/seconds

Explanation:

The velocity of an object is the rate of change of the displacement of the object with time measured in m/s. It is similar to speed except that it has a direction and is thus a vector.

Mathematically,

Velocity = Displacement/time

Given that the ball drops 450 m in 2 seconds

Velocity of the ball = 450 m/ 2 seconds

= 225 m/sec

Answer 2

Explanation:

If the distance fallen after seconds is denoted by s(t) and measured in meters, then Galileo's law is expressed by the equation s(t)=
`4.9t^2`.

Average velocity=
`(change in position)/(Time elapsed)`

=
`(s(2.1)-s(2))/(0.1)`

=
`(4.9(2.1)^2-4.9(2)^2)/(0.1)`

=
`(4.9(4.41)-4.9(4))/(0.1)`

=
`(21.60-19.6)/(0.1)`

=
`(2)/(0.1)`

=
`20ms^(-1)`

Thus, the velocity of the ball after 2 seconds will be
`20ms^(-1)`.