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Answer 1 · 2015-11-26T01:11:40

$33.34"m/s"^2$

Once the stone is released we consider the vertical component of its motion to get the time of flight.

Always try and get $t$ first because this is common to both vertical and horizontal components of the journey.

$s=1/2"g"t^2$

$:.t=sqrt((2s)/(g))$

$:.t=sqrt((2xxcancel(10))/(cancel(10))$

$t=1.414"s"$

The horizontal component of the stone's velocity is constant so we can get it from:

$v=s/t=10/1.414=7.07"m/s"$

The centripetal acceleration of the stone at the point of release is given by:

$a=(v^2)/r=(7.07^2)/1.5=33.34"m/s"^2$