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If a ball is thrown into the air with a velocity of $40 \mathrm{ft} / \mathrm{s},$ itsheight (in feet) after $t$ seconds is given by $y=40 t-16 t^{2}$ .Find the velocity when $t=2$

$$v(2)=-24 \text { feet } \mathrm{s}$$

Calculus 1 / AB

Chapter 2

DERIVATIVES

Section 1

Derivatives and Rates of Change

Derivatives

Differentiation

Applications of the Derivative

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the problem. You see Equation 40 team in the 16 T Square as well as the initial velocity of 40 feet per second. The problem asked to find the velocity AT T was too. Which basically means finding the isn't thing is really changed. So we plugged in the equation. Uh, institutions may change for one TZ with two we'll get be of two is equal to the limit this t goes to of effort T minus effort to over T minus two. Now you have the final one of two is equal to. So if you're putting two into the equation, OK, 40 terms to 16 this equal to a T minus 64 which gives you 16. You plug it back into the equation limit t to you. Plug in the equation. 40 T minus 16 t squared, minus 16. No routine rightist, too. And now you factor out the negative eight to help simplify when there is a TV host to negative p times to t square five t close to oh, teen. Now you factor out to t square minus five t plus two a woman t ghost to he to t minus one SWAT team minus two. All their teams to and I can cancel up t minus two. This gets you a little. It has t goes with to open native eight to t minus once. And now, since there's nothing else, you've been simplified. What? You to just plug in the to the equation. So too minus one. Then this gets you negative. Eight times. Four minutes? Well, just three. Get you native. 24. He cursed sick.

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