Question

An arrow is shot at an initial velocity of 224 ft/sec from a height of 44 feet, ay what time will the arrow be 824 feet above the ground?

Answer 1

The arrow is subject to the constant downward pull of gravity, so its vertical position over time is given by

$y=44\,\mathrm{ft}+\left(224\,\dfrac{\mathrm{ft}}{\mathrm s}\right)t+\dfrac12\left(-32.1\,\dfrac{\mathrm{ft}}{\mathrm s^2}\right)t^2$

We want to find $t$ for which $y=824\,\mathrm{ft}$. You should find two solutions at $t=6.67\,\mathrm s$ and $t=7.29\,\mathrm s$.