Rates of Changes

Problem 149: An object is dropped from the top of a 100-m-high tower. Its height above the ground after $t$ sec is $100 - 4.9t^2$ m. How fast is it falling 2 sec after it is dropped?

Solution: We will let $y$ be the height above the ground after $t$ sec. That is, $y = 100 - 4.9t^2$ . Since we are interested on how fast the object is falling, we can use derivative (average rate of change) to find our solution.

(1) $\begin{align*} y(t) & = 100 - 4.9t^2 \\ y'(t) & = -9.8t \\ y'(2 ) & = -9.8(2) \quad \text{(at $t=2$)} \\ & = -19.6 \ \text{m/s}. \end{align*}$

Hence, the object is falling at a speed of 19.6 m/s.

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