#### Answer

Between 1 s and 2 s, the projectile's height exceeds $32$ feet.

#### Work Step by Step

We are given that $ s(t)=-16t^2+v_0t +s_0$
Set $ v_0=48$
Thus, $0=-16(0)^2+(48)(0) +s_0$
so, $ s_0=0$
Now, $-16t^2+48t=-32 \implies t^2-3t+2=0$
This implies that $(t-2)(t-1)=0$
or, $ t= 1 s, 2 s $
Therefore, between 1 s and 2 s, the projectile's height exceeds $32$ feet.