Answer
$(a)\space -9.8\space m/s^{2}$
$(b)\space 5.7\space m$
Work Step by Step
(a) The pebble is subjected to acceleration due to gravity once it has left the slingshot. Since the downward direction is negative, the acceleration of the pebble is $=-9.8\space m/s^{2} $ and the magnitude of the pebble's downward velocity is increasing, not decreasing.
(b)Here we use the equation 2.8 $S=ut+\frac{1}{2}at^{2}$ to find the pebbles' displacement.
$\downarrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$S=9\space m/s\times0.5\space s+\frac{1}{2}\times9.8\space m/s^{2}\times (0.5\space s)^{2}=5.7\space m$
So, after 0.5 s, the pebble is 5.7 m beneath the cliff-top
