# Question #fe1a9

As the ball is thrown horizontally with velocity $u = 12 \frac{m}{s}$ its initial vertical component will be zero and the time taken for free fall of height, $h = 28 m$ under gravity will be its time (t) required to land.
So $h = \frac{1}{2} \cdot g \cdot {t}^{2}$
$\implies t = \sqrt{\frac{2 h}{g}} = \frac{\sqrt{2 \cdot 28}}{9.8} \approx 2.39 s$
$S = u \times t = 12 \times 2.39 m \approx 28.68 m$