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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. 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

User Coldy
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Answer:

v = -19.6 m / s, y =y₀ + v₀ t - ½ g t²

Step-by-step explanation:

This is an exercise in kinetics in one dimension, let's take the upward direction as positive

v = v₀ - g t

in this case as the body is released its initial velocity is zero and the acceleration is -g the sign indicates that it is directed downwards

v = 0 -g t

v = - 9.8 2

v = -19.6 m / s

the sign indicates that the speed is down.

Galileo's equation is

y =y₀ + v₀ t - ½ g t²

where i is the initial height, v₀ the initial velocity and -g the acceleration of the body

User Josef Borkovec
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