# Solving limits with free fall

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• Jan 21st 2013, 07:54 AM
minneola24
Solving limits with free fall
Hello,

Just started calculus. I am fairly comfortable with limit problems in what has been given to us, but I'm thoroughly lost when given a problem like this.

Any help would be appreciated, thanks.
• Jan 21st 2013, 09:34 AM
hollywood
Re: Solving limits with free fall
When you replace s(a) and s(t) with $\displaystyle -4.9a^2+200$ and $\displaystyle -4.9t^2+200$, you get a lot of cancellation. Remember the identity $\displaystyle x^2-y^2=(x+y)(x-y)$.

They're getting you ready for derivatives - the limit you are calculating is just the derivative of s(t) at a.

- Hollywood
• Jan 22nd 2013, 06:42 AM
minneola24
Re: Solving limits with free fall
So s(a) and s(t) are the exact same? Why do they have different letters then?
• Jan 22nd 2013, 06:57 AM
hollywood
Re: Solving limits with free fall
No, $\displaystyle s(a)=-4.9a^2+200$ and $\displaystyle s(t)=-4.9t^2+200$. So when you plug them in, 200 cancels and you can factor out -4.9, giving $\displaystyle \frac{a^2-t^2}{a-t}$.

- Hollywood