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Answer 1 · 2016-10-12T04:36:07

As the ball is thrown horizontally with velocity $u = 12 \frac{m}{s}$ $u=12m/s$ its initial vertical component will be zero and the time taken for free fall of height, $h = 28 m$ $h= 28m$ under gravity will be its time (t) required to land.

So $h = \frac{1}{2} \cdot g \cdot t^{2}$ $h=1/2*g*t^2$
$\Rightarrow t = \sqrt{\frac{2 h}{g}} = \frac{\sqrt{2 \cdot 28}}{9.8} \approx 2.39 s$ $=>t=sqrt((2h)/g)=sqrt(2*28)/9.8~~2.39s$

During this time of fall the horizontal component will remain undiminished,supposing no resistance of air.So the horizontal diatance traversed during this time of fall will be

$S = u \times t = 12 \times 2.39 m \approx 28.68 m$ $S=uxxt=12xx2.39m~~28.68m$

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