Question

Question #a2a1d

Answer 1

`33.34"m/s"^2`

Explanation:

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`