# How long does a ball take to reach the ground 7 m below, if it is thrown straight up with an initial speed of 2.00 m/s?

Oct 29, 2014

u=2m/s

a= -g = -10m/${s}^{2}$

S= -7m

t= ?

$S = u t + \frac{1}{2} a {t}^{2}$
$- 7 = 2 t + \frac{1}{2} \left(- 10\right) {t}^{2}$
$2 t - 5 {t}^{2} + 7 = 0$
$5 {t}^{2} - 2 t - 7 = 0$
$t = \frac{7}{5} , - 1$

As t can't be negative, t = 7/5 s = 1.4s.

Therefore, t = 1.4s