3. A rock is thrown from the top of a tall
building. The distance, in feet, between the
rock and the ground t seconds after it is
thrown is given by
d(t) =
4t + 382.
- 16t²
a. How long after the rock is thrown is it
370 feet from the ground?
b. After how many seconds will the rock
reach the ground?
c. What is the height of the rock when it
is halfway to the ground?

where d is the distance (in feet) between the rock and ground t seconds after thrown. Ground is 0 feet. As a note, I'm not clear why the "4t" term is in this equation. It has the effect of adding an increase in height at the rate of 4 feet/sec. I'll assume this is to accommodate any increase in height due to the throw, but it is a bit curious.

a. How long after the rock is thrown is it 370 feet from the ground?

The 382 in the equation represents the starting height of the rock (382 feet). To find the time to reach 370 feet:

370 = 4t + 382 - 16t²

Solve using the quadratic equation. There will be two answers, one negative, which we'll discard until we can go backwards in time. The positive value is 1, which would mean the rock will be at 370 feet in 1 second after throwing.

b. After how many seconds will the rock reach the ground?

Ground is 0 feet:

0 = 4t + 382 - 16t²

t = 5 seconds

c. What is the height of the rock when it is halfway to the ground?

The initial height is 382 feet, as per the equation, since when t = o (initial), d(t) is 382 feet. Halfway to the ground would be (382/2) or 191 feet.