# A coin is dropped from a height of 750 feet. The height, *s* (ft), at time *t* (sec), is given by s=-16t^2+750. How long does it take for the coin to hit the ground?

## & what is the velocity when the coin hits the ground?

Nov 1, 2016

$6.85$sec

#### Explanation:

At ground level, $s = 0$

$s = 0 \implies - 16 {t}^{2} + 750 = 0$
$\therefore 16 {t}^{2} = 750$
$\therefore {t}^{2} = \frac{750}{16}$
$\therefore {t}^{2} = \frac{375}{8}$
$\therefore t = \pm \sqrt{\frac{375}{8}}$
$\therefore t = \pm 6.85$ (2dp)

We need $t > 0$, and so $t = 6.85$s (2dp)