Answer
$0.75m/s^2$
Work Step by Step
Step 1. Given $s(t)=15t-\frac{1}{2}g_st^2$, we can find the velocity as $v(t)=s'(t)=15-g_st$ m/s.
Step 2. At $t=20$ sec, the ball bearing reached its maximum height, which means that $v(20)=0$ and this gives $15-20g_s=0$; thus we get $g_s=15/20=3/4=0.75m/s^2$
