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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#