# Question a2a1d

Nov 26, 2015

$33.34 {\text{m/s}}^{2}$

#### 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.

$s = \frac{1}{2} \text{g} {t}^{2}$

$\therefore t = \sqrt{\frac{2 s}{g}}$

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

$t = 1.414 \text{s}$

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

$v = \frac{s}{t} = \frac{10}{1.414} = 7.07 \text{m/s}$

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

$a = \frac{{v}^{2}}{r} = \frac{{7.07}^{2}}{1.5} = 33.34 {\text{m/s}}^{2}$