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\,\frac{\mathrm{ft}}{\mathrm s}\right)t+\frac12\left(-32.1\,\frac{\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`.