Question

An object dropped from a height falls a distance of \(d\) feet in \(t\) seconds. The formula that describes this relationship is \(d=16 t^{2} .\) A math book is dropped from the top of a building that is 400 feet high. a. How far has the book fallen in \(\frac{1}{2}\) second? b. After \(\frac{3}{4}\) seconds, how far above the ground is the book?

Short answer

Answer: After 0.5 seconds, the book has fallen 4 feet. After 3/4 seconds, the book has fallen 9 feet. After 3/4 seconds, the book is 391 feet above the ground.

Step-by-step solution

Calculate the distance dropped after 0.5 seconds

To find out how far the book has fallen in 0.5 seconds, we plug \(t = \frac{1}{2}\) into the given formula, \(d=16 t^{2}\): \[ d=16 \times \left(\frac{1}{2}\right)^{2} \]

Evaluate the Expression

Now, substitute the values and calculate the distance: \[ d = 16 \times \left(\frac{1}{4}\right) = 16 \times 0.25 = 4 \] Therefore, after \(\frac{1}{2}\) second, the book has fallen 4 feet.

Calculate the distance dropped after 3/4 seconds

To find out how far the book has fallen in \(\frac{3}{4}\) seconds, we plug t = 3/4 into the given formula, \(d=16 t^{2}\): \[ d=16 \times \left(\frac{3}{4}\right)^{2} \]

Evaluate the Expression

Now, substitute the values and calculate the distance: \[ d = 16 \times \left(\frac{9}{16}\right) = 9 \] Therefore, after \(\frac{3}{4}\) seconds, the book has fallen 9 feet.

Calculate the remaining distance to the ground

To find how far above the ground the book is after \(\frac{3}{4}\) seconds, we subtract the distance fallen (9 feet) from the total height of the building (400 feet): \[ \text{Remaining distance} = 400 - 9 = 391 \text{ feet} \] So, after \(\frac{3}{4}\) seconds, the book is 391 feet above the ground.