Instantaneous Rate of Change of Height

[Calc12]

A soccor ball is kicked into the air. The polynomial function f(t) = -4.9^2 +16t + 1, where the f(t) represents the height in metres at t seconds models this scenario.
Determine the instantaneous rate of change of height at 1 second, 2 seconds and 3 seconds.

Im thinking I use this:

f ' (1) = lim f(1 + h) - f(1)
______h?0____h


and so forth for 2 seconds, and 3 seconds..



But I got 87.71 m/s for 1 second - doesn't seem right?



thanks in advance

 

[soroban]

Hello, Calc12!

You left out a "square" in the function . . .

A soccor ball is kicked into the air.
\(\displaystyle \text{The polynomial function: }\:f(t) \:=\: -4.9t^2 +16t + 1\,\text{ represents the height in meters at }t\text{ seconds.}\)

Determine the instantaneous rate of change of height at 1 second, 2 seconds and 3 seconds.

\(\displaystyle \text{You want to use: }\:f'(t) \;=\;\lim_{h\to0}\frac{f(t+h) - f(t)}{h}\)

. . \(\displaystyle \text{And you should get: }\:f'(t) \;=\;-9.8t + 16\)


\(\displaystyle \text{Then evaluate: }\:f'(1),\;f'(2),\;f'(3).\)

 

[Calc12]

Thanks again Soroban.. You don't even understand how much of a saviour you and this site truly is!