Solutions

a) We have $s''(t)=-32$ so $s'(t)=-32t+\alpha$. Hence,

$$s(t)=-16t^2+\alpha t+\beta.$$

b) Clearly $\alpha=v_0$ and $s_0=\beta$.

c) In this case,

$$s(t)=-16t^2+vt.$$

Since it is a polynomial of even degree with a negative leading coefficient, it attains a global maximum when $s'(t)=0$. This corresponds to

$$t=\frac{v}{32}.$$