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?

Question

asked Feb 23, 2019 199k views

1 Answer

Emiguevara · Answer 1 · 2019-03-02T00:33:57+0000

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