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.

Respuesta :

Answer:

Step-by-step 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)=[tex]4.9t^2[/tex].

Average velocity=[tex]\frac{change in position}{Time elapsed}[/tex]

=[tex]\frac{s(2.1)-s(2)}{0.1}[/tex]

=[tex]\frac{4.9(2.1)^2-4.9(2)^2}{0.1}[/tex]

=[tex]\frac{4.9(4.41)-4.9(4)}{0.1}[/tex]

=[tex]\frac{21.60-19.6}{0.1}[/tex]

=[tex]\frac{2}{0.1}[/tex]

=[tex]20ms^{-1}[/tex]

Thus, the velocity of the ball after 2 seconds will be [tex]20ms^{-1}[/tex].

Answer:

225 m/seconds

Step-by-step 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