# What's the difficulty in finding instantaneous velocity?

EXAMPLE 1 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 5 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 seconds is denoted by and measured in meters, then Galileo’s law is expressed by the equation

$$s(t)= 4.9t^2$$

The difficulty in finding the velocity after 5 s is that we are dealing with a single instant of time $(t=5)$ , 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 $t=5$ to $t=5.1$:

What he meant with difficulty here?

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Isn't instantaneous velocity just $v_{avg}=\frac{\Delta d}{\Delta t}=\frac{d_f-d_i}{t_f-t_i}$? – Voyska Sep 30 '12 at 2:51
That fraction only makes sense if $t_f\neq t_i$, but that means that you're not looking at an instant. – user22805 Sep 30 '12 at 2:56
It may be a very dumb question, but why can't we just $t$ for $5$ in $s(t)= 4.9t^2$? – Voyska Sep 30 '12 at 23:28
Here, $s(t)$ stands for displacement (that is, the distance fallen), not velocity or speed. – user22805 Sep 30 '12 at 23:48