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Answer 1 · 2017-05-12T05:12:57

$122.5 m$ from the top of the tower.

Using $v^2-u^2=at$ , to find the velocity of the ball at the $5th$ second of the fall.

Given $u=0$ , $a=9.8$ , $t=5$
we get $v= 49 m*s^-1$

Distance traveled by the stone is

$s=vt-(1/2)(g)(t^2)$

$s=49*5-(1/2)*10*5^2$

$s=122.5$

The stone is at distance of $122.5 m$ from the top of the tower at the $5th$ second.